{"id":179,"date":"2026-01-12T16:00:39","date_gmt":"2026-01-12T16:00:39","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/?post_type=chapter&#038;p=179"},"modified":"2026-01-12T16:00:39","modified_gmt":"2026-01-12T16:00:39","slug":"advanced-integration-techniques-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/advanced-integration-techniques-get-stronger-answer-key\/","title":{"raw":"Advanced Integration Techniques: Get Stronger Answer Key","rendered":"Advanced Integration Techniques: Get Stronger Answer Key"},"content":{"raw":"<h2>Integration by Parts<\/h2>\r\n<ol>\r\n \t<li>[latex]u={x}^{3}[\/latex]<\/li>\r\n \t<li>[latex]u={y}^{3}[\/latex]<\/li>\r\n \t<li>[latex]u=\\sin\\left(2x\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\text{-}x+x\\text{ln}x+C[\/latex]<\/li>\r\n \t<li>[latex]x{\\tan}^{-1}x-\\frac{1}{2}\\text{ln}\\left(1+{x}^{2}\\right)+C[\/latex]<\/li>\r\n \t<li>[latex]-\\frac{1}{2}x\\cos\\left(2x\\right)+\\frac{1}{4}\\sin\\left(2x\\right)+C[\/latex]<\/li>\r\n \t<li>[latex]{e}^{\\text{-}x}\\left(-1-x\\right)+C[\/latex]<\/li>\r\n \t<li>[latex]2x\\cos{x}+\\left(-2+{x}^{2}\\right)\\sin{x}+C[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2}\\left(1+2x\\right)\\left(-1+\\text{ln}\\left(1+2x\\right)\\right)+C[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2}{e}^{x}\\left(\\text{-}\\cos{x}+\\sin{x}\\right)+C[\/latex]<\/li>\r\n \t<li>[latex]-\\frac{{e}^{\\text{-}{x}^{2}}}{2}+C[\/latex]<\/li>\r\n \t<li>[latex]-\\frac{1}{2}x\\cos\\left[\\text{ln}\\left(2x\\right)\\right]+\\frac{1}{2}x\\sin\\left[\\text{ln}\\left(2x\\right)\\right]+C[\/latex]<\/li>\r\n \t<li>[latex]2x - 2x\\text{ln}x+x{\\left(\\text{ln}x\\right)}^{2}+C[\/latex]<\/li>\r\n \t<li>[latex]\\left(\\text{-}\\frac{{x}^{3}}{9}+\\frac{1}{3}{x}^{3}\\text{ln}x\\right)+C[\/latex]<\/li>\r\n \t<li>[latex]-\\frac{1}{2}\\sqrt{1 - 4{x}^{2}}+x{\\cos}^{-1}\\left(2x\\right)+C[\/latex]<\/li>\r\n \t<li>[latex]\\text{-}\\left(-2+{x}^{2}\\right)\\cos{x}+2x\\sin{x}+C[\/latex]<\/li>\r\n \t<li>[latex]\\text{-}x\\left(-6+{x}^{2}\\right)\\cos{x}+3\\left(-2+{x}^{2}\\right)\\sin{x}+C[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2}x\\left(\\text{-}\\sqrt{1-\\frac{1}{{x}^{2}}}+x\\cdot {\\sec}^{-1}x\\right)+C[\/latex]<\/li>\r\n \t<li>[latex]\\text{-}\\text{cosh}x+x\\text{sinh}x+C[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{4}-\\frac{3}{4{\\text{e}}^{2}}[\/latex]<\/li>\r\n \t<li>[latex]2[\/latex]<\/li>\r\n \t<li>[latex]2\\pi [\/latex]<\/li>\r\n \t<li>[latex]-2+\\pi [\/latex]<\/li>\r\n \t<li>[latex]\\text{-}\\sin\\left(x\\right)+\\text{ln}\\left[\\sin\\left(x\\right)\\right]\\sin{x}+C[\/latex]<\/li>\r\n \t<li>Answers vary\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\frac{2}{5}\\left(1+x\\right){\\left(-3+2x\\right)}^{\\frac{3}{2}}+C[\/latex]<\/li>\r\n \t<li>[latex]\\frac{2}{5}\\left(1+x\\right){\\left(-3+2x\\right)}^{\\frac{3}{2}}+C[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Do not use integration by parts. Choose [latex]u[\/latex] to be [latex]\\text{ln}x[\/latex], and the integral is of the form [latex]\\displaystyle\\int {u}^{2}du[\/latex].<\/li>\r\n \t<li>Do not use integration by parts. Let [latex]u={x}^{2}-3[\/latex], and the integral can be put into the form [latex]\\displaystyle\\int {e}^{u}du[\/latex].<\/li>\r\n \t<li>Do not use integration by parts. Choose [latex]u[\/latex] to be [latex]u=3{x}^{3}+2[\/latex] and the integral can be put into the form [latex]\\displaystyle\\int \\sin\\left(u\\right)du[\/latex].<\/li>\r\n \t<li>\r\n<p id=\"fs-id1165042271819\">The area under graph is [latex]0.39535[\/latex].<span data-type=\"newline\">\r\n<\/span><\/p>\r\n<span id=\"fs-id1165042271823\" data-type=\"media\" data-alt=\"This figure is the graph of y=e^-x sin(pi*x). The curve begins in the third quadrant at x=0.5, increases through the origin, reaches a high point between 0.5 and 0.75, then decreases, passing through x=1.\">\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233939\/CNX_Calc_Figure_07_01_202.jpg\" alt=\"This figure is the graph of y=e^-x sin(pi*x). The curve begins in the third quadrant at x=0.5, increases through the origin, reaches a high point between 0.5 and 0.75, then decreases, passing through x=1.\" data-media-type=\"image\/jpeg\" \/><\/span><\/li>\r\n \t<li>[latex]2\\pi e[\/latex]<\/li>\r\n \t<li>[latex]2.05[\/latex]<\/li>\r\n \t<li>[latex]12\\pi [\/latex]<\/li>\r\n \t<li>[latex]8{\\pi }^{2}[\/latex]<\/li>\r\n<\/ol>\r\n<h2>Trigonometric Integrals<\/h2>\r\n<ol>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{{\\sin}^{4}x}{4}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{12}{\\tan}^{6}\\left(2x\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{\\sec}^{2}\\left(\\dfrac{x}{2}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{-}\\cos{x}+\\dfrac{1}{3}\\cos^{2}x+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{2}{\\cos}^{2}x+C[\/latex] or [latex]\\dfrac{1}{2}{\\sin}^{2}x+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{3}\\cos^{3}x+\\dfrac{2}{5}\\cos^{5}x-\\dfrac{1}{7}\\cos^{7}x+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2}{3}{\\left(\\sin{x}\\right)}^{\\dfrac{3}{2}}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sec{x}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\sec{x}\\tan{x}-\\dfrac{1}{2}\\text{ln}\\left(\\sec{x}+\\tan{x}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2\\tan{x}}{3}+\\dfrac{1}{3}\\sec{\\left(x\\right)}^{2}\\tan{x}[\/latex] [latex]=\\tan{x}+\\dfrac{{\\tan}^{3}x}{3}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{-}\\text{ln}|\\cot{x}+\\csc{x}|+C[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\sin}^{3}\\left(ax\\right)}{3a}+C[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{\\pi }{2}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{x}{2}+\\dfrac{1}{12}\\sin\\left(6x\\right)+C[\/latex]<\/li>\r\n \t<li>[latex]x+C[\/latex]<\/li>\r\n \t<li>[latex]0[\/latex]<\/li>\r\n \t<li>[latex]0[\/latex]<\/li>\r\n \t<li>[latex]0[\/latex]<\/li>\r\n \t<li>[latex]\\text{Approximately 0.239}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{2}[\/latex]<\/li>\r\n \t<li>[latex]1.0[\/latex]<\/li>\r\n \t<li>[latex]0[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{3\\theta }{8}-\\dfrac{1}{4\\pi }\\sin\\left(2\\pi \\theta \\right)+\\dfrac{1}{32\\pi }\\sin\\left(4\\pi \\theta \\right)+C=f\\left(x\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\text{ln}\\left(\\sqrt{3}\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\displaystyle\\int }_{\\text{-}\\pi }^{\\pi }\\sin\\left(2x\\right)\\cos\\left(3x\\right)dx=0[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{\\tan\\left(x\\right)}x\\left(\\dfrac{8\\tan{x}}{21}+\\dfrac{2}{7}\\sec{x}^{2}\\tan{x}\\right)+C=f\\left(x\\right)[\/latex]<\/li>\r\n \t<li>The second integral is more difficult because the first integral is simply a u-substitution type.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]9{\\tan}^{2}\\theta [\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{a}^{2}{\\text{cosh}}^{2}\\theta [\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]4{\\left(x-\\dfrac{1}{2}\\right)}^{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{-}{\\left(x+1\\right)}^{2}+5[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}|x+\\sqrt{\\text{-}{a}^{2}+{x}^{2}}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{3}\\text{ln}|\\sqrt{9{x}^{2}+1}+3x|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{\\sqrt{1-{x}^{2}}}{x}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]9\\left[\\dfrac{x\\sqrt{{x}^{2}+9}}{18}+\\dfrac{1}{2}ln|\\dfrac{\\sqrt{{x}^{2}+9}}{3}+\\dfrac{x}{3}|\\right]+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{3}\\sqrt{9-{\\theta }^{2}}\\left(18+{\\theta }^{2}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{\\sqrt{1+{x}^{2}}}{x}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{8}\\left(x\\left(5 - 2{x}^{2}\\right)\\sqrt{1-{x}^{2}}+3\\text{arcsin}x\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}x-\\text{ln}|1+\\sqrt{1-{x}^{2}}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{\\sqrt{-1+{x}^{2}}}{x}+\\text{ln}|x+\\sqrt{-1+{x}^{2}}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{\\sqrt{1+{x}^{2}}}{x}+\\text{arcsinh}x+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{1+x}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2\\sqrt{-10+x}\\sqrt{x}\\text{ln}|\\sqrt{-10+x}+\\sqrt{x}|}{\\sqrt{\\left(10-x\\right)x}}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{9\\pi }{2}[\/latex]; area of a semicircle with radius 3<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{arcsin}\\left(x\\right)+C[\/latex] is the common answer.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\text{ln}\\left(1+{x}^{2}\\right)+C[\/latex] is the result using either method.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use trigonometric substitution. Let [latex]x=\\sec\\left(\\theta \\right)[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]4.367[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{{\\pi }^{2}}{8}+\\dfrac{\\pi }{4}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]y=\\dfrac{1}{16}\\text{ln}|\\dfrac{x+8}{x - 8}|+3[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]24.6[\/latex] m\u00b3<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2\\pi }{3}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<h2>Partial Fractions<\/h2>\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{2}{x+1}+\\dfrac{5}{2\\left(x+2\\right)}+\\dfrac{1}{2x}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{{x}^{2}}+\\dfrac{3}{x}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]2{x}^{2}+4x+8+\\dfrac{16}{x - 2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{{x}^{2}}-\\dfrac{1}{x}+\\dfrac{1}{x - 1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{2\\left(x - 2\\right)}+\\dfrac{1}{2\\left(x - 1\\right)}-\\dfrac{1}{6x}+\\dfrac{1}{6\\left(x - 3\\right)}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{x - 1}+\\dfrac{2x+1}{{x}^{2}+x+1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2}{x+1}+\\dfrac{x}{{x}^{2}+4}-\\dfrac{1}{{\\left({x}^{2}+4\\right)}^{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{-}\\text{ln}|2-x|+2\\text{ln}|4+x|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\text{ln}|4-{x}^{2}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]2\\left(x+\\dfrac{1}{3}\\text{arctan}\\left(\\dfrac{1+x}{3}\\right)\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]2\\text{ln}|x|-3\\text{ln}|1+x|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{16}\\left(\\text{-}\\dfrac{4}{-2+x}-\\text{ln}|-2+x|+\\text{ln}|2+x|\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{30}\\left(-2\\sqrt{5}\\text{arctan}\\left[\\dfrac{1+x}{\\sqrt{5}}\\right]+2\\text{ln}|-4+x|-\\text{ln}|6+2x+{x}^{2}|\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{3}{x}+4\\text{ln}|x+2|+x+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{-}\\text{ln}|3-x|+\\dfrac{1}{2}\\text{ln}|{x}^{2}+4|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}|x - 2|-\\dfrac{1}{2}\\text{ln}|{x}^{2}+2x+2|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{-}x+\\text{ln}|1-{e}^{x}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{5}\\text{ln}|\\dfrac{\\cos{x}+3}{\\cos{x} - 2}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2 - 2{e}^{2t}}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]2\\sqrt{1+x}-2\\text{ln}|1+\\sqrt{1+x}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}|\\dfrac{\\sin{x}}{1-\\sin{x}}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\sqrt{3}}{4}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x-\\text{ln}\\left(1+{e}^{x}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]6{x}^{\\dfrac{1}{6}}-3{x}^{\\dfrac{1}{3}}+2\\sqrt{x}-6\\text{ln}\\left(1+{x}^{\\dfrac{1}{6}}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{4}{3}\\pi \\text{arctanh}\\left[\\dfrac{1}{3}\\right]=\\dfrac{1}{3}\\pi \\text{ln}4[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x=\\text{-}\\text{ln}|t - 3|+\\text{ln}|t - 4|+\\text{ln}2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x=\\text{ln}|t - 1|-\\sqrt{2}\\text{arctan}\\left(\\sqrt{2}t\\right)-\\dfrac{1}{2}\\text{ln}\\left({t}^{2}+\\dfrac{1}{2}\\right)+\\sqrt{2}\\text{arctan}\\left(2\\sqrt{2}\\right)+\\dfrac{1}{2}\\text{ln}4.5[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2}{5}\\pi \\text{ln}\\dfrac{28}{13}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\text{arctan}\\left[\\dfrac{-1+2x}{\\sqrt{3}}\\right]}{\\sqrt{3}}+\\dfrac{1}{3}\\text{ln}|1+x|-\\dfrac{1}{6}\\text{ln}|1-x+{x}^{2}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]2.0[\/latex] in.\u00b2<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]3{\\left(-8+x\\right)}^{\\dfrac{1}{3}}[\/latex] [latex]-2\\sqrt{3}\\text{arctan}\\left[\\dfrac{-1+{\\left(-8+x\\right)}^{\\dfrac{1}{3}}}{\\sqrt{3}}\\right][\/latex] [latex]-2\\text{ln}\\left[2+{\\left(-8+x\\right)}^{\\dfrac{1}{3}}\\right][\/latex] [latex]+\\text{ln}\\left[4 - 2{\\left(-8+x\\right)}^{\\dfrac{1}{3}}+{\\left(-8+x\\right)}^{\\dfrac{2}{3}}\\right]+C[\/latex]<\/li>\r\n<\/ol>\r\n<h2>Other Strategies for Integration<\/h2>\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\text{ln}|{x}^{2}+2x+2|+2\\text{arctan}\\left(x+1\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{\\text{cosh}}^{-1}\\left(\\dfrac{x+3}{3}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{{2}^{{x}^{2}-1}}{\\text{ln}2}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{arcsin}\\left(\\dfrac{y}{2}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{2}\\csc\\left(2w\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]9 - 6\\sqrt{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]2-\\dfrac{\\pi }{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{12}{\\tan}^{4}\\left(3x\\right)-\\dfrac{1}{6}{\\tan}^{2}\\left(3x\\right)+\\dfrac{1}{3}\\text{ln}|\\sec\\left(3x\\right)|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]2\\cot\\left(\\dfrac{w}{2}\\right)-2\\csc\\left(\\dfrac{w}{2}\\right)+w+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{5}\\text{ln}|\\dfrac{2\\left(5+4\\sin{t} - 3\\cos{t}\\right)}{4\\cos{t}+3\\sin{t}}|[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]6{x}^{\\dfrac{1}{6}}-3{x}^{\\dfrac{1}{3}}+2\\sqrt{x}-6\\text{ln}\\left[1+{x}^{\\dfrac{1}{6}}\\right]+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{-}{x}^{3}\\cos{x}+3{x}^{2}\\sin{x}+6x\\cos{x} - 6\\sin{x}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\left({x}^{2}+\\text{ln}|1+{e}^{\\text{-}{x}^{2}}|\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]2\\text{arctan}\\left(\\sqrt{x - 1}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]0.5=\\dfrac{1}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]8.0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{3}\\text{arctan}\\left(\\dfrac{1}{3}\\left(x+2\\right)\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{3}\\text{arctan}\\left(\\dfrac{x+1}{3}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}\\left({e}^{x}+\\sqrt{4+{e}^{2x}}\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}x-\\dfrac{1}{6}\\text{ln}\\left({x}^{6}+1\\right)-\\dfrac{\\text{arctan}\\left({x}^{3}\\right)}{3{x}^{3}}+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}|x+\\sqrt{16+{x}^{2}}|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{4}\\cot\\left(2x\\right)+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\text{arctan}10[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]1276.14[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]7.21[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sqrt{5}-\\sqrt{2}+\\text{ln}|\\dfrac{2+2\\sqrt{2}}{1+\\sqrt{5}}|[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{3}\\text{arctan}\\left(3\\right)\\approx 0.416[\/latex]<\/li>\r\n<\/ol>","rendered":"<h2>Integration by Parts<\/h2>\n<ol>\n<li>[latex]u={x}^{3}[\/latex]<\/li>\n<li>[latex]u={y}^{3}[\/latex]<\/li>\n<li>[latex]u=\\sin\\left(2x\\right)[\/latex]<\/li>\n<li>[latex]\\text{-}x+x\\text{ln}x+C[\/latex]<\/li>\n<li>[latex]x{\\tan}^{-1}x-\\frac{1}{2}\\text{ln}\\left(1+{x}^{2}\\right)+C[\/latex]<\/li>\n<li>[latex]-\\frac{1}{2}x\\cos\\left(2x\\right)+\\frac{1}{4}\\sin\\left(2x\\right)+C[\/latex]<\/li>\n<li>[latex]{e}^{\\text{-}x}\\left(-1-x\\right)+C[\/latex]<\/li>\n<li>[latex]2x\\cos{x}+\\left(-2+{x}^{2}\\right)\\sin{x}+C[\/latex]<\/li>\n<li>[latex]\\frac{1}{2}\\left(1+2x\\right)\\left(-1+\\text{ln}\\left(1+2x\\right)\\right)+C[\/latex]<\/li>\n<li>[latex]\\frac{1}{2}{e}^{x}\\left(\\text{-}\\cos{x}+\\sin{x}\\right)+C[\/latex]<\/li>\n<li>[latex]-\\frac{{e}^{\\text{-}{x}^{2}}}{2}+C[\/latex]<\/li>\n<li>[latex]-\\frac{1}{2}x\\cos\\left[\\text{ln}\\left(2x\\right)\\right]+\\frac{1}{2}x\\sin\\left[\\text{ln}\\left(2x\\right)\\right]+C[\/latex]<\/li>\n<li>[latex]2x - 2x\\text{ln}x+x{\\left(\\text{ln}x\\right)}^{2}+C[\/latex]<\/li>\n<li>[latex]\\left(\\text{-}\\frac{{x}^{3}}{9}+\\frac{1}{3}{x}^{3}\\text{ln}x\\right)+C[\/latex]<\/li>\n<li>[latex]-\\frac{1}{2}\\sqrt{1 - 4{x}^{2}}+x{\\cos}^{-1}\\left(2x\\right)+C[\/latex]<\/li>\n<li>[latex]\\text{-}\\left(-2+{x}^{2}\\right)\\cos{x}+2x\\sin{x}+C[\/latex]<\/li>\n<li>[latex]\\text{-}x\\left(-6+{x}^{2}\\right)\\cos{x}+3\\left(-2+{x}^{2}\\right)\\sin{x}+C[\/latex]<\/li>\n<li>[latex]\\frac{1}{2}x\\left(\\text{-}\\sqrt{1-\\frac{1}{{x}^{2}}}+x\\cdot {\\sec}^{-1}x\\right)+C[\/latex]<\/li>\n<li>[latex]\\text{-}\\text{cosh}x+x\\text{sinh}x+C[\/latex]<\/li>\n<li>[latex]\\frac{1}{4}-\\frac{3}{4{\\text{e}}^{2}}[\/latex]<\/li>\n<li>[latex]2[\/latex]<\/li>\n<li>[latex]2\\pi[\/latex]<\/li>\n<li>[latex]-2+\\pi[\/latex]<\/li>\n<li>[latex]\\text{-}\\sin\\left(x\\right)+\\text{ln}\\left[\\sin\\left(x\\right)\\right]\\sin{x}+C[\/latex]<\/li>\n<li>Answers vary\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{2}{5}\\left(1+x\\right){\\left(-3+2x\\right)}^{\\frac{3}{2}}+C[\/latex]<\/li>\n<li>[latex]\\frac{2}{5}\\left(1+x\\right){\\left(-3+2x\\right)}^{\\frac{3}{2}}+C[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Do not use integration by parts. Choose [latex]u[\/latex] to be [latex]\\text{ln}x[\/latex], and the integral is of the form [latex]\\displaystyle\\int {u}^{2}du[\/latex].<\/li>\n<li>Do not use integration by parts. Let [latex]u={x}^{2}-3[\/latex], and the integral can be put into the form [latex]\\displaystyle\\int {e}^{u}du[\/latex].<\/li>\n<li>Do not use integration by parts. Choose [latex]u[\/latex] to be [latex]u=3{x}^{3}+2[\/latex] and the integral can be put into the form [latex]\\displaystyle\\int \\sin\\left(u\\right)du[\/latex].<\/li>\n<li>\n<p id=\"fs-id1165042271819\">The area under graph is [latex]0.39535[\/latex].<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p><span id=\"fs-id1165042271823\" data-type=\"media\" data-alt=\"This figure is the graph of y=e^-x sin(pi*x). The curve begins in the third quadrant at x=0.5, increases through the origin, reaches a high point between 0.5 and 0.75, then decreases, passing through x=1.\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233939\/CNX_Calc_Figure_07_01_202.jpg\" alt=\"This figure is the graph of y=e^-x sin(pi*x). The curve begins in the third quadrant at x=0.5, increases through the origin, reaches a high point between 0.5 and 0.75, then decreases, passing through x=1.\" data-media-type=\"image\/jpeg\" \/><\/span><\/li>\n<li>[latex]2\\pi e[\/latex]<\/li>\n<li>[latex]2.05[\/latex]<\/li>\n<li>[latex]12\\pi[\/latex]<\/li>\n<li>[latex]8{\\pi }^{2}[\/latex]<\/li>\n<\/ol>\n<h2>Trigonometric Integrals<\/h2>\n<ol>\n<li style=\"list-style-type: none;\">\n<ol>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{{\\sin}^{4}x}{4}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{12}{\\tan}^{6}\\left(2x\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{\\sec}^{2}\\left(\\dfrac{x}{2}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{-}\\cos{x}+\\dfrac{1}{3}\\cos^{2}x+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{2}{\\cos}^{2}x+C[\/latex] or [latex]\\dfrac{1}{2}{\\sin}^{2}x+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{3}\\cos^{3}x+\\dfrac{2}{5}\\cos^{5}x-\\dfrac{1}{7}\\cos^{7}x+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2}{3}{\\left(\\sin{x}\\right)}^{\\dfrac{3}{2}}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\sec{x}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\sec{x}\\tan{x}-\\dfrac{1}{2}\\text{ln}\\left(\\sec{x}+\\tan{x}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2\\tan{x}}{3}+\\dfrac{1}{3}\\sec{\\left(x\\right)}^{2}\\tan{x}[\/latex] [latex]=\\tan{x}+\\dfrac{{\\tan}^{3}x}{3}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{-}\\text{ln}|\\cot{x}+\\csc{x}|+C[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\sin}^{3}\\left(ax\\right)}{3a}+C[\/latex]<\/li>\n<li>[latex]\\dfrac{\\pi }{2}[\/latex]<\/li>\n<li>[latex]\\dfrac{x}{2}+\\dfrac{1}{12}\\sin\\left(6x\\right)+C[\/latex]<\/li>\n<li>[latex]x+C[\/latex]<\/li>\n<li>[latex]0[\/latex]<\/li>\n<li>[latex]0[\/latex]<\/li>\n<li>[latex]0[\/latex]<\/li>\n<li>[latex]\\text{Approximately 0.239}[\/latex]<\/li>\n<li>[latex]\\sqrt{2}[\/latex]<\/li>\n<li>[latex]1.0[\/latex]<\/li>\n<li>[latex]0[\/latex]<\/li>\n<li>[latex]\\dfrac{3\\theta }{8}-\\dfrac{1}{4\\pi }\\sin\\left(2\\pi \\theta \\right)+\\dfrac{1}{32\\pi }\\sin\\left(4\\pi \\theta \\right)+C=f\\left(x\\right)[\/latex]<\/li>\n<li>[latex]\\text{ln}\\left(\\sqrt{3}\\right)[\/latex]<\/li>\n<li>[latex]{\\displaystyle\\int }_{\\text{-}\\pi }^{\\pi }\\sin\\left(2x\\right)\\cos\\left(3x\\right)dx=0[\/latex]<\/li>\n<li>[latex]\\sqrt{\\tan\\left(x\\right)}x\\left(\\dfrac{8\\tan{x}}{21}+\\dfrac{2}{7}\\sec{x}^{2}\\tan{x}\\right)+C=f\\left(x\\right)[\/latex]<\/li>\n<li>The second integral is more difficult because the first integral is simply a u-substitution type.<\/li>\n<li class=\"whitespace-normal break-words\">[latex]9{\\tan}^{2}\\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{a}^{2}{\\text{cosh}}^{2}\\theta[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]4{\\left(x-\\dfrac{1}{2}\\right)}^{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{-}{\\left(x+1\\right)}^{2}+5[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}|x+\\sqrt{\\text{-}{a}^{2}+{x}^{2}}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{3}\\text{ln}|\\sqrt{9{x}^{2}+1}+3x|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{\\sqrt{1-{x}^{2}}}{x}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]9\\left[\\dfrac{x\\sqrt{{x}^{2}+9}}{18}+\\dfrac{1}{2}ln|\\dfrac{\\sqrt{{x}^{2}+9}}{3}+\\dfrac{x}{3}|\\right]+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{3}\\sqrt{9-{\\theta }^{2}}\\left(18+{\\theta }^{2}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{\\sqrt{1+{x}^{2}}}{x}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{8}\\left(x\\left(5 - 2{x}^{2}\\right)\\sqrt{1-{x}^{2}}+3\\text{arcsin}x\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}x-\\text{ln}|1+\\sqrt{1-{x}^{2}}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{\\sqrt{-1+{x}^{2}}}{x}+\\text{ln}|x+\\sqrt{-1+{x}^{2}}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{\\sqrt{1+{x}^{2}}}{x}+\\text{arcsinh}x+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{1+x}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2\\sqrt{-10+x}\\sqrt{x}\\text{ln}|\\sqrt{-10+x}+\\sqrt{x}|}{\\sqrt{\\left(10-x\\right)x}}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{9\\pi }{2}[\/latex]; area of a semicircle with radius 3<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{arcsin}\\left(x\\right)+C[\/latex] is the common answer.<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\text{ln}\\left(1+{x}^{2}\\right)+C[\/latex] is the result using either method.<\/li>\n<li class=\"whitespace-normal break-words\">Use trigonometric substitution. Let [latex]x=\\sec\\left(\\theta \\right)[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">[latex]4.367[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{{\\pi }^{2}}{8}+\\dfrac{\\pi }{4}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]y=\\dfrac{1}{16}\\text{ln}|\\dfrac{x+8}{x - 8}|+3[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]24.6[\/latex] m\u00b3<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2\\pi }{3}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h2>Partial Fractions<\/h2>\n<ol>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{2}{x+1}+\\dfrac{5}{2\\left(x+2\\right)}+\\dfrac{1}{2x}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{{x}^{2}}+\\dfrac{3}{x}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]2{x}^{2}+4x+8+\\dfrac{16}{x - 2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{{x}^{2}}-\\dfrac{1}{x}+\\dfrac{1}{x - 1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{2\\left(x - 2\\right)}+\\dfrac{1}{2\\left(x - 1\\right)}-\\dfrac{1}{6x}+\\dfrac{1}{6\\left(x - 3\\right)}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{x - 1}+\\dfrac{2x+1}{{x}^{2}+x+1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2}{x+1}+\\dfrac{x}{{x}^{2}+4}-\\dfrac{1}{{\\left({x}^{2}+4\\right)}^{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{-}\\text{ln}|2-x|+2\\text{ln}|4+x|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\text{ln}|4-{x}^{2}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]2\\left(x+\\dfrac{1}{3}\\text{arctan}\\left(\\dfrac{1+x}{3}\\right)\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]2\\text{ln}|x|-3\\text{ln}|1+x|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{16}\\left(\\text{-}\\dfrac{4}{-2+x}-\\text{ln}|-2+x|+\\text{ln}|2+x|\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{30}\\left(-2\\sqrt{5}\\text{arctan}\\left[\\dfrac{1+x}{\\sqrt{5}}\\right]+2\\text{ln}|-4+x|-\\text{ln}|6+2x+{x}^{2}|\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{3}{x}+4\\text{ln}|x+2|+x+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{-}\\text{ln}|3-x|+\\dfrac{1}{2}\\text{ln}|{x}^{2}+4|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}|x - 2|-\\dfrac{1}{2}\\text{ln}|{x}^{2}+2x+2|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{-}x+\\text{ln}|1-{e}^{x}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{5}\\text{ln}|\\dfrac{\\cos{x}+3}{\\cos{x} - 2}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2 - 2{e}^{2t}}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]2\\sqrt{1+x}-2\\text{ln}|1+\\sqrt{1+x}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}|\\dfrac{\\sin{x}}{1-\\sin{x}}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\sqrt{3}}{4}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x-\\text{ln}\\left(1+{e}^{x}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]6{x}^{\\dfrac{1}{6}}-3{x}^{\\dfrac{1}{3}}+2\\sqrt{x}-6\\text{ln}\\left(1+{x}^{\\dfrac{1}{6}}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{4}{3}\\pi \\text{arctanh}\\left[\\dfrac{1}{3}\\right]=\\dfrac{1}{3}\\pi \\text{ln}4[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x=\\text{-}\\text{ln}|t - 3|+\\text{ln}|t - 4|+\\text{ln}2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x=\\text{ln}|t - 1|-\\sqrt{2}\\text{arctan}\\left(\\sqrt{2}t\\right)-\\dfrac{1}{2}\\text{ln}\\left({t}^{2}+\\dfrac{1}{2}\\right)+\\sqrt{2}\\text{arctan}\\left(2\\sqrt{2}\\right)+\\dfrac{1}{2}\\text{ln}4.5[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{2}{5}\\pi \\text{ln}\\dfrac{28}{13}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{\\text{arctan}\\left[\\dfrac{-1+2x}{\\sqrt{3}}\\right]}{\\sqrt{3}}+\\dfrac{1}{3}\\text{ln}|1+x|-\\dfrac{1}{6}\\text{ln}|1-x+{x}^{2}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]2.0[\/latex] in.\u00b2<\/li>\n<li class=\"whitespace-normal break-words\">[latex]3{\\left(-8+x\\right)}^{\\dfrac{1}{3}}[\/latex] [latex]-2\\sqrt{3}\\text{arctan}\\left[\\dfrac{-1+{\\left(-8+x\\right)}^{\\dfrac{1}{3}}}{\\sqrt{3}}\\right][\/latex] [latex]-2\\text{ln}\\left[2+{\\left(-8+x\\right)}^{\\dfrac{1}{3}}\\right][\/latex] [latex]+\\text{ln}\\left[4 - 2{\\left(-8+x\\right)}^{\\dfrac{1}{3}}+{\\left(-8+x\\right)}^{\\dfrac{2}{3}}\\right]+C[\/latex]<\/li>\n<\/ol>\n<h2>Other Strategies for Integration<\/h2>\n<ol>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\text{ln}|{x}^{2}+2x+2|+2\\text{arctan}\\left(x+1\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{\\text{cosh}}^{-1}\\left(\\dfrac{x+3}{3}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{{2}^{{x}^{2}-1}}{\\text{ln}2}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{arcsin}\\left(\\dfrac{y}{2}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{2}\\csc\\left(2w\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]9 - 6\\sqrt{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]2-\\dfrac{\\pi }{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{12}{\\tan}^{4}\\left(3x\\right)-\\dfrac{1}{6}{\\tan}^{2}\\left(3x\\right)+\\dfrac{1}{3}\\text{ln}|\\sec\\left(3x\\right)|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]2\\cot\\left(\\dfrac{w}{2}\\right)-2\\csc\\left(\\dfrac{w}{2}\\right)+w+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{5}\\text{ln}|\\dfrac{2\\left(5+4\\sin{t} - 3\\cos{t}\\right)}{4\\cos{t}+3\\sin{t}}|[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]6{x}^{\\dfrac{1}{6}}-3{x}^{\\dfrac{1}{3}}+2\\sqrt{x}-6\\text{ln}\\left[1+{x}^{\\dfrac{1}{6}}\\right]+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{-}{x}^{3}\\cos{x}+3{x}^{2}\\sin{x}+6x\\cos{x} - 6\\sin{x}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\left({x}^{2}+\\text{ln}|1+{e}^{\\text{-}{x}^{2}}|\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]2\\text{arctan}\\left(\\sqrt{x - 1}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]0.5=\\dfrac{1}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]8.0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{3}\\text{arctan}\\left(\\dfrac{1}{3}\\left(x+2\\right)\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{3}\\text{arctan}\\left(\\dfrac{x+1}{3}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}\\left({e}^{x}+\\sqrt{4+{e}^{2x}}\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}x-\\dfrac{1}{6}\\text{ln}\\left({x}^{6}+1\\right)-\\dfrac{\\text{arctan}\\left({x}^{3}\\right)}{3{x}^{3}}+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}|x+\\sqrt{16+{x}^{2}}|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]-\\dfrac{1}{4}\\cot\\left(2x\\right)+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{2}\\text{arctan}10[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]1276.14[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]7.21[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\sqrt{5}-\\sqrt{2}+\\text{ln}|\\dfrac{2+2\\sqrt{2}}{1+\\sqrt{5}}|[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{3}\\text{arctan}\\left(3\\right)\\approx 0.416[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":15,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":109,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/179"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":1,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/179\/revisions"}],"predecessor-version":[{"id":191,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/179\/revisions\/191"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/109"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/179\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=179"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=179"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=179"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=179"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}