{"id":689,"date":"2025-06-20T17:03:54","date_gmt":"2025-06-20T17:03:54","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=689"},"modified":"2025-09-10T13:59:55","modified_gmt":"2025-09-10T13:59:55","slug":"integration-by-parts-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-by-parts-learn-it-2\/","title":{"raw":"Integration by Parts: Learn It 2","rendered":"Integration by Parts: Learn It 2"},"content":{"raw":"

How to Choose [latex]u[\/latex] and[latex] dv[\/latex]<\/h2>\r\nThe natural question you're probably asking is: How do I know which part should be [latex]u[\/latex] and which should be [latex]dv[\/latex]?<\/em> Sometimes it takes some trial and error, but there's a helpful strategy that can guide your choices.\r\n\r\n
\r\n

The LIATE Method<\/strong><\/p>\r\n

The acronym LIATE<\/strong> can help take the guesswork out of choosing [latex]u[\/latex] and [latex]dv[\/latex]:<\/p>\r\n\r\n

    \r\n \t
  • L<\/strong>ogarithmic Functions<\/li>\r\n \t
  • I<\/strong>nverse Trigonometric Functions<\/li>\r\n \t
  • A<\/strong>lgebraic Functions<\/li>\r\n \t
  • T<\/strong>rigonometric Functions<\/li>\r\n \t
  • E<\/strong>xponential Functions<\/li>\r\n<\/ul>\r\n

    Rule:<\/strong> Choose [latex]u[\/latex] to be the function type that appears first<\/strong> in this list.<\/p>\r\n\r\n<\/section>For example, if your integral contains both a logarithmic function and an algebraic function, choose [latex]u[\/latex] to be the logarithmic function since L<\/strong> comes before A<\/strong> in LIATE.\r\n

    Why Does LIATE Work?<\/h3>\r\n

    This mnemonic works because of integration practicality:<\/p>\r\n\r\n

      \r\n \t
    • Logarithmic and inverse trig functions<\/strong> are at the front because we don't have simple integration formulas for them\u2014so they make poor choices for [latex]dv[\/latex]<\/li>\r\n \t
    • Exponential and trig functions<\/strong> are at the end because they're easy to integrate and make excellent choices for [latex]dv[\/latex]<\/li>\r\n \t
    • Algebraic functions<\/strong> are in the middle because they're generally manageable both to integrate and differentiate<\/li>\r\n<\/ul>\r\n
      To use the by-parts technique successfully, it is helpful to first review the derivative rules of several familiar transcendental functions.\r\n
        \r\n \t
      1. [latex] \\frac{d}{dx} (\\sin x) = \\cos x [\/latex]<\/li>\r\n \t
      2. [latex] \\frac{d}{dx} (\\cos x) = -\\sin x [\/latex]<\/li>\r\n \t
      3. [latex] \\frac{d}{dx} (\\ln x) = \\frac{1}{x} [\/latex]<\/li>\r\n \t
      4. [latex] \\frac{d}{dx} (\\arcsin x) = \\frac{1}{\\sqrt{1-x^2}} [\/latex]<\/li>\r\n \t
      5. [latex] \\frac{d}{dx} (\\arctan x) = \\frac{1}{1+x^2} [\/latex]<\/li>\r\n<\/ol>\r\n<\/section>Now let's see how this method works in practice.\r\n\r\n
        \r\n
        \r\n

        Evaluate [latex]\\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n

        \r\n

        Begin by rewriting the integral:<\/p>\r\n\r\n

        [latex]\\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx={\\displaystyle\\int }^{\\text{ }}{x}^{-3}\\text{ln}xdx[\/latex].<\/div>\r\n

        Since this integral contains the algebraic function [latex]{x}^{-3}[\/latex] and the logarithmic function [latex]\\text{ln}x[\/latex], choose [latex]u=\\text{ln}x[\/latex], since L comes before A in LIATE. After we have chosen [latex]u=\\text{ln}x[\/latex], we must choose [latex]dv={x}^{-3}dx[\/latex].<\/p>\r\n

        Next, since [latex]u=\\text{ln}x[\/latex], we have [latex]du=\\frac{1}{x}dx[\/latex]. Also, [latex]v={\\displaystyle\\int }^{\\text{ }}{x}^{-3}dx=-\\frac{1}{2}{x}^{-2}[\/latex]. Summarizing,<\/p>\r\n\r\n

        [latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & \\text{ln}x\\hfill & & \\hfill dv& =\\hfill & {x}^{-3}dx\\hfill \\\\ \\hfill du& =\\hfill & \\frac{1}{x}dx\\hfill & & \\hfill v& =\\hfill & {\\displaystyle\\int }^{\\text{ }}{x}^{-3}dx=-\\frac{1}{2}{x}^{-2}.\\end{array}[\/latex]<\/div>\r\n

        Substituting into the integration-by-parts formula gives<\/p>\r\n\r\n

        [latex]\\begin{array}{ccccc}\\hfill \\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx& ={\\displaystyle\\int }^{\\text{ }}{x}^{-3}\\text{ln}xdx=\\left(\\text{ln}x\\right)\\left(\\text{-}\\frac{1}{2}{x}^{-2}\\right)-{\\displaystyle\\int }^{\\text{ }}\\left(\\text{-}\\frac{1}{2}{x}^{-2}\\right)\\left(\\frac{1}{x}dx\\right)\\hfill & & & \\\\ & =-\\frac{1}{2}{x}^{-2}\\text{ln}x+{\\displaystyle\\int }^{\\text{ }}\\frac{1}{2}{x}^{-3}dx\\hfill & & & \\text{Simplify}.\\hfill \\\\ & =-\\frac{1}{2}{x}^{-2}\\text{ln}x-\\frac{1}{4}{x}^{-2}+C\\hfill & & & \\text{Integrate}.\\hfill \\\\ & =-\\frac{1}{2{x}^{2}}\\text{ln}x-\\frac{1}{4{x}^{2}}+C.\\hfill & & & \\text{Rewrite with positive integers.}\\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>In some cases it may be necessary to apply integration by parts more than once.\r\n\r\n
        \r\n
        \r\n

        Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n

        \r\n

        Using LIATE, choose [latex]u={x}^{2}[\/latex] and [latex]dv={e}^{3x}dx[\/latex]. Thus, [latex]du=2xdx[\/latex] and [latex]v=\\displaystyle\\int {e}^{3x}dx=\\left(\\frac{1}{3}\\right){e}^{3x}[\/latex]. Therefore,<\/p>\r\n\r\n

        [latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & {x}^{2}\\hfill & & \\hfill dv& =\\hfill & {e}^{3x}dx\\hfill \\\\ \\hfill du& =\\hfill & 2xdx\\hfill & & \\hfill v& =\\hfill & \\displaystyle\\int {e}^{3x}dx=\\frac{1}{3}{e}^{3x}.\\end{array}[\/latex]<\/div>\r\n \r\n

        Substituting into the integration-by-parts formula\u00a0produces<\/p>\r\n\r\n

        [latex]\\displaystyle\\int {x}^{2}{e}^{3x}dx=\\frac{1}{3}{x}^{2}{e}^{3x}-\\displaystyle\\int \\frac{2}{3}x{e}^{3x}dx[\/latex].<\/div>\r\n \r\n

        We still cannot integrate [latex]\\displaystyle\\int \\frac{2}{3}x{e}^{3x}dx[\/latex] directly, but the integral now has a lower power on [latex]x[\/latex]. We can evaluate this new integral by using integration by parts again. To do this, choose [latex]u=x[\/latex] and [latex]dv=\\frac{2}{3}{e}^{3x}dx[\/latex]. Thus, [latex]du=dx[\/latex] and [latex]v=\\displaystyle\\int \\left(\\frac{2}{3}\\right){e}^{3x}dx=\\left(\\frac{2}{9}\\right){e}^{3x}[\/latex]. Now we have<\/p>\r\n\r\n

        [latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & x\\hfill & & \\hfill dv& =\\hfill & \\frac{2}{3}{e}^{3x}dx\\hfill \\\\ \\hfill du& =\\hfill & dx\\hfill & & \\hfill v& =\\hfill & \\displaystyle\\int \\frac{2}{3}{e}^{3x}dx=\\frac{2}{9}{e}^{3x}.\\end{array}[\/latex]<\/div>\r\n \r\n

        Substituting back into the previous equation yields<\/p>\r\n\r\n

        [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx=\\frac{1}{3}{x}^{2}{e}^{3x}-\\left(\\frac{2}{9}x{e}^{3x}-{\\displaystyle\\int }^{\\text{ }}\\frac{2}{9}{e}^{3x}dx\\right)[\/latex].<\/div>\r\n \r\n

        After evaluating the last integral and simplifying, we obtain<\/p>\r\n\r\n

        [latex]\\displaystyle\\int {x}^{2}{e}^{3x}dx=\\frac{1}{3}{x}^{2}{e}^{3x}-\\frac{2}{9}x{e}^{3x}+\\frac{2}{27}{e}^{3x}+C[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
        \r\n
        \r\n

        Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n\r\nThis integral appears to have only one function\u2014namely, [latex]\\sin\\left(\\text{ln}x\\right)[\/latex] \u2014however, we can always use the constant function 1 as the other function. In this example, let\u2019s choose [latex]u=\\sin\\left(\\text{ln}x\\right)[\/latex] and [latex]dv=1dx[\/latex]. (The decision to use [latex]u=\\sin\\left(\\text{ln}x\\right)[\/latex] is easy. We can\u2019t choose [latex]dv=\\sin\\left(\\text{ln}x\\right)dx[\/latex] because if we could integrate it, we wouldn\u2019t be using integration by parts in the first place!) Consequently, [latex]du=\\frac{1}{x}\\cos\\left(\\text{ln}x\\right)dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}1dx=x[\/latex]. After applying integration by parts to the integral and simplifying, we have\r\n

        \r\n
        [latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=x\\sin\\left(\\text{ln}x\\right)-{\\displaystyle\\int }^{\\text{ }}\\cos\\left(\\text{ln}x\\right)dx[\/latex].<\/div>\r\n \r\n

        Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let\u2019s see what happens when we apply integration by parts again. This time let\u2019s choose [latex]u=\\cos\\left(\\text{ln}x\\right)[\/latex] and [latex]dv=1dx[\/latex], making [latex]du=\\text{-}\\frac{1}{x}\\sin\\left(\\text{ln}x\\right)dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}1dx=x[\/latex]. Substituting, we have<\/p>\r\n\r\n

        [latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=x\\sin\\left(\\text{ln}x\\right)-\\left(x\\cos\\left(\\text{ln}x\\right)-{\\displaystyle\\int }^{\\text{ }}-\\sin\\left(\\text{ln}x\\right)dx\\right)[\/latex].<\/div>\r\n \r\n

        After simplifying, we obtain<\/p>\r\n\r\n

        [latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=x\\sin\\left(\\text{ln}x\\right)-x\\cos\\left(\\text{ln}x\\right)-{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex].<\/div>\r\n \r\n

        The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we can actually evaluate the integral. To see how to do this more clearly, substitute [latex]I={\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex]. Thus, the equation becomes<\/p>\r\n\r\n

        [latex]I=x\\sin\\left(\\text{ln}x\\right)-x\\cos\\left(\\text{ln}x\\right)-I[\/latex].<\/div>\r\n \r\n

        First, add [latex]I[\/latex] to both sides of the equation to obtain<\/p>\r\n\r\n

        [latex]2I=x\\sin\\left(\\text{ln}x\\right)-x\\cos\\left(\\text{ln}x\\right)[\/latex].<\/div>\r\n \r\n

        Next, divide by 2:<\/p>\r\n\r\n

        [latex]I=\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)[\/latex].<\/div>\r\n \r\n

        Substituting [latex]I={\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx[\/latex] again, we have<\/p>\r\n\r\n

        [latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)[\/latex].<\/div>\r\n \r\n

        From this we see that [latex]\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)[\/latex] is an antiderivative of [latex]\\sin\\left(\\text{ln}x\\right)dx[\/latex]. For the most general antiderivative, add [latex]+C\\text{:}[\/latex]<\/p>\r\n\r\n

        [latex]{\\displaystyle\\int }^{\\text{ }}\\sin\\left(\\text{ln}x\\right)dx=\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)+C[\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n
        \r\n
        Analysis<\/strong><\/div>\r\n

        If this method feels a little strange at first, we can check the answer by differentiation:<\/p>\r\n\r\n

        [latex]\\begin{array}{c}\\frac{d}{dx}\\left(\\frac{1}{2}x\\sin\\left(\\text{ln}x\\right)-\\frac{1}{2}x\\cos\\left(\\text{ln}x\\right)\\right)\\hfill \\\\ \\\\ =\\frac{1}{2}\\left(\\sin\\left(\\text{ln}x\\right)\\right)+\\cos\\left(\\text{ln}x\\right)\\cdot \\frac{1}{x}\\cdot \\frac{1}{2}x-\\left(\\frac{1}{2}\\cos\\left(\\text{ln}x\\right)-\\sin\\left(\\text{ln}x\\right)\\cdot \\frac{1}{x}\\cdot \\frac{1}{2}x\\right)\\hfill \\\\ =\\sin\\left(\\text{ln}x\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
        \"Caution\"Caution! LIATE is a guide, not a rigid rule. If your first choice leads to an integral you can't evaluate, try a different approach.<\/section>
        \r\n
        \r\n

        Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{t}^{3}{e}^{{t}^{2}}dt[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n

        \r\n

        If we use a strict interpretation of the mnemonic LIATE to make our choice of [latex]u[\/latex], we end up with [latex]u={t}^{3}[\/latex] and [latex]dv={e}^{{t}^{2}}dt[\/latex]. Unfortunately, this choice won\u2019t work because we are unable to evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{e}^{{t}^{2}}dt[\/latex]. However, since we can evaluate [latex]{\\displaystyle\\int }^{\\text{ }}t{e}^{{t}^{2}}dx[\/latex], we can try choosing [latex]u={t}^{2}[\/latex] and [latex]dv=t{e}^{{t}^{2}}dt[\/latex]. With these choices we have<\/p>\r\n\r\n

        [latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & {t}^{2}\\hfill & & \\hfill dv& =\\hfill & t{e}^{{t}^{2}}dt\\hfill \\\\ \\hfill du& =\\hfill & 2tdt\\hfill & & \\hfill v& =\\hfill & {\\displaystyle\\int }^{\\text{ }}t{e}^{{t}^{2}}dt=\\frac{1}{2}{e}^{{t}^{2}}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

        Thus, we obtain<\/p>\r\n\r\n

        [latex]\\begin{array}{cc}{\\displaystyle\\int {t}^{3}{e}^{{t}^{2}}dt}\\hfill & =\\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-{\\displaystyle\\int \\frac{1}{2}{e}^{{t}^{2}}2tdt}\\hfill \\\\ \\hfill & =\\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-\\frac{1}{2}{e}^{{t}^{2}}+C.\\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
        [ohm_question hide_question_numbers=1]311291[\/ohm_question]<\/section>","rendered":"

        How to Choose [latex]u[\/latex] and[latex]dv[\/latex]<\/h2>\n

        The natural question you’re probably asking is: How do I know which part should be [latex]u[\/latex] and which should be [latex]dv[\/latex]?<\/em> Sometimes it takes some trial and error, but there’s a helpful strategy that can guide your choices.<\/p>\n

        \n

        The LIATE Method<\/strong><\/p>\n

        The acronym LIATE<\/strong> can help take the guesswork out of choosing [latex]u[\/latex] and [latex]dv[\/latex]:<\/p>\n

          \n
        • L<\/strong>ogarithmic Functions<\/li>\n
        • I<\/strong>nverse Trigonometric Functions<\/li>\n
        • A<\/strong>lgebraic Functions<\/li>\n
        • T<\/strong>rigonometric Functions<\/li>\n
        • E<\/strong>xponential Functions<\/li>\n<\/ul>\n

          Rule:<\/strong> Choose [latex]u[\/latex] to be the function type that appears first<\/strong> in this list.<\/p>\n<\/section>\n

          For example, if your integral contains both a logarithmic function and an algebraic function, choose [latex]u[\/latex] to be the logarithmic function since L<\/strong> comes before A<\/strong> in LIATE.<\/p>\n

          Why Does LIATE Work?<\/h3>\n

          This mnemonic works because of integration practicality:<\/p>\n

            \n
          • Logarithmic and inverse trig functions<\/strong> are at the front because we don’t have simple integration formulas for them\u2014so they make poor choices for [latex]dv[\/latex]<\/li>\n
          • Exponential and trig functions<\/strong> are at the end because they’re easy to integrate and make excellent choices for [latex]dv[\/latex]<\/li>\n
          • Algebraic functions<\/strong> are in the middle because they’re generally manageable both to integrate and differentiate<\/li>\n<\/ul>\n
            To use the by-parts technique successfully, it is helpful to first review the derivative rules of several familiar transcendental functions.<\/p>\n
              \n
            1. [latex]\\frac{d}{dx} (\\sin x) = \\cos x[\/latex]<\/li>\n
            2. [latex]\\frac{d}{dx} (\\cos x) = -\\sin x[\/latex]<\/li>\n
            3. [latex]\\frac{d}{dx} (\\ln x) = \\frac{1}{x}[\/latex]<\/li>\n
            4. [latex]\\frac{d}{dx} (\\arcsin x) = \\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/li>\n
            5. [latex]\\frac{d}{dx} (\\arctan x) = \\frac{1}{1+x^2}[\/latex]<\/li>\n<\/ol>\n<\/section>\n

              Now let’s see how this method works in practice.<\/p>\n

              \n
              \n

              Evaluate [latex]\\displaystyle\\int \\frac{\\text{ln}x}{{x}^{3}}dx[\/latex].<\/p>\n<\/div>\n