{"id":409,"date":"2025-02-13T19:44:41","date_gmt":"2025-02-13T19:44:41","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-involving-exponential-and-logarithmic-functions-learn-it-2\/"},"modified":"2025-02-13T19:44:41","modified_gmt":"2025-02-13T19:44:41","slug":"integrals-involving-exponential-and-logarithmic-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-involving-exponential-and-logarithmic-functions-learn-it-2\/","title":{"raw":"Integrals Involving Exponential and Logarithmic Functions: Learn It 2","rendered":"Integrals Involving Exponential and Logarithmic Functions: Learn It 2"},"content":{"raw":"\n<h2>Integrals&nbsp;Involving Logarithmic Functions<\/h2>\n<p id=\"fs-id1170572565333\">Integrating functions of the form [latex]f(x)={x}^{-1}[\/latex] result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as [latex]f(x)=\\text{ln}x[\/latex] and [latex]f(x)={\\text{log}}_{a}x,[\/latex] are also included in the rule.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>integration formulas involving logarithmic functions<\/h3>\n<p id=\"fs-id1170572543663\">The following formulas can be used to evaluate integrals involving logarithmic functions.<\/p>\n<div id=\"fs-id1170572543666\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\displaystyle\\int {x}^{-1}dx&amp; =\\hfill &amp; \\text{ln}|x|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\text{ln}xdx&amp; =\\hfill &amp; x\\text{ln}x-x+C=x(\\text{ln}x-1)+C\\hfill \\\\ \\hfill \\displaystyle\\int {\\text{log}}_{a}xdx&amp; =\\hfill &amp; \\frac{x}{\\text{ln}a}(\\text{ln}x-1)+C\\hfill \\end{array}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of the function [latex]\\dfrac{3}{x-10}.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1170571653435\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571653435\"]\n\n<p id=\"fs-id1170571653435\">First factor the [latex]3[\/latex] outside the integral symbol. Then use the [latex]u^{-1}[\/latex]&nbsp;rule. Thus,<\/p>\n<div id=\"fs-id1170571653444\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\displaystyle\\int \\frac{3}{x-10}dx}\\hfill &amp; =3{\\displaystyle\\int \\frac{1}{x-10}dx}\\hfill \\\\ &amp; =3{\\displaystyle\\int \\frac{du}{u}}\\hfill \\\\ &amp; =3\\text{ln}|u|+C\\hfill \\\\ &amp; =3\\text{ln}|x-10|+C,x\\ne 10.\\hfill \\end{array}[\/latex]<\/div>\n\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204256\/CNX_Calc_Figure_05_06_003.jpg\" alt=\"A graph of the function f(x) = 3 \/ (x \u2013 10). There is an asymptote at x=10. The first segment is a decreasing concave down curve that approaches 0 as x goes to negative infinity and approaches negative infinity as x goes to 10. The second segment is a decreasing concave up curve that approaches infinity as x goes to 10 and approaches 0 as x approaches infinity.\" width=\"325\" height=\"246\"> Figure 3. The domain of this function is [latex]x\\ne 10.[\/latex][\/caption]\n[\/hidden-answer]<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of [latex]\\dfrac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1170572480517\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572480517\"]<br>\nThis can be rewritten as [latex]\\displaystyle\\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx.[\/latex] <br>\n<br>\nUse substitution. Let [latex]u={x}^{4}+3{x}^{2},[\/latex] then [latex]du=4{x}^{3}+6x.[\/latex] Alter <em>du<\/em> by factoring out the [latex]2[\/latex]. <br>\n<br>\nThus,\n\n<div id=\"fs-id1170571807881\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\hfill du&amp; =\\hfill &amp; (4{x}^{3}+6x)dx\\hfill \\\\ &amp; =\\hfill &amp; 2(2{x}^{3}+3x)dx\\hfill \\\\ \\hfill \\frac{1}{2}du&amp; =\\hfill &amp; (2{x}^{3}+3x)dx.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572373464\">Rewrite the integrand in [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170572331833\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx=\\frac{1}{2}\\displaystyle\\int {u}^{-1}du.[\/latex]<\/div>\n<p id=\"fs-id1170571733965\">Then we have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}\\frac{1}{2}{\\displaystyle\\int {u}^{-1}du}\\hfill &amp; =\\frac{1}{2}\\text{ln}|u|+C\\hfill \\\\ &amp; =\\frac{1}{2}\\text{ln}|{x}^{4}+3{x}^{2}|+C.\\hfill \\end{array}[\/latex]<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8mW2bG6HoPE?controls=0&amp;start=162&amp;end=270&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.6.2_162to270_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.6.2\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of the log function [latex]{\\text{log}}_{2}x.[\/latex]<br>\n[reveal-answer q=\"fs-id1170571649954\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571649954\"]<\/p>\n<p id=\"fs-id1170571649954\">Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have<\/p>\n<div id=\"fs-id1170571649959\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {\\text{log}}_{2}xdx=\\frac{x}{\\text{ln}2}(\\text{ln}x-1)+C.[\/latex][\/hidden-answer]<\/div>\n<\/section>\n<p>The example below is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.<\/p>\n<section class=\"textbox example\">\n<p>Find the definite integral of [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\frac{ \\sin x}{1+ \\cos x}dx.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1170571636148\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170571636148\"]\n\n<p id=\"fs-id1170571636148\">We need substitution to evaluate this problem. Let [latex]u=1+ \\cos x,,[\/latex] so [latex]du=\\text{\u2212} \\sin xdx.[\/latex] Rewrite the integral in terms of [latex]u[\/latex], changing the limits of integration as well. Thus,<\/p>\n<div id=\"fs-id1170571636203\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}u=1+ \\cos (0)=2\\hfill \\\\ u=1+ \\cos (\\frac{\\pi }{2})=1\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572510108\">Then<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int} _{0}^{\\pi \\text{\/}2}\\dfrac{ \\sin x}{1+ \\cos x}\\hfill &amp; =\\text{\u2212}{\\displaystyle\\int }_{2}^{1}{u}^{-1}du\\hfill&nbsp; \\\\ &amp; ={\\displaystyle\\int }_{1}^{2}{u}^{-1}du\\hfill \\\\ &amp; ={\\text{ln}|u||}_{1}^{2}\\hfill \\\\ &amp; =\\left[\\text{ln}2-\\text{ln}1\\right]\\hfill \\\\ &amp; =\\text{ln}2\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]288437[\/ohm_question]<\/p>\n<\/section>\n","rendered":"<h2>Integrals&nbsp;Involving Logarithmic Functions<\/h2>\n<p id=\"fs-id1170572565333\">Integrating functions of the form [latex]f(x)={x}^{-1}[\/latex] result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as [latex]f(x)=\\text{ln}x[\/latex] and [latex]f(x)={\\text{log}}_{a}x,[\/latex] are also included in the rule.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>integration formulas involving logarithmic functions<\/h3>\n<p id=\"fs-id1170572543663\">The following formulas can be used to evaluate integrals involving logarithmic functions.<\/p>\n<div id=\"fs-id1170572543666\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\displaystyle\\int {x}^{-1}dx& =\\hfill & \\text{ln}|x|+C\\hfill \\\\ \\hfill \\displaystyle\\int \\text{ln}xdx& =\\hfill & x\\text{ln}x-x+C=x(\\text{ln}x-1)+C\\hfill \\\\ \\hfill \\displaystyle\\int {\\text{log}}_{a}xdx& =\\hfill & \\frac{x}{\\text{ln}a}(\\text{ln}x-1)+C\\hfill \\end{array}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of the function [latex]\\dfrac{3}{x-10}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571653435\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571653435\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571653435\">First factor the [latex]3[\/latex] outside the integral symbol. Then use the [latex]u^{-1}[\/latex]&nbsp;rule. Thus,<\/p>\n<div id=\"fs-id1170571653444\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\displaystyle\\int \\frac{3}{x-10}dx}\\hfill & =3{\\displaystyle\\int \\frac{1}{x-10}dx}\\hfill \\\\ & =3{\\displaystyle\\int \\frac{du}{u}}\\hfill \\\\ & =3\\text{ln}|u|+C\\hfill \\\\ & =3\\text{ln}|x-10|+C,x\\ne 10.\\hfill \\end{array}[\/latex]<\/div>\n<figure style=\"width: 325px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204256\/CNX_Calc_Figure_05_06_003.jpg\" alt=\"A graph of the function f(x) = 3 \/ (x \u2013 10). There is an asymptote at x=10. The first segment is a decreasing concave down curve that approaches 0 as x goes to negative infinity and approaches negative infinity as x goes to 10. The second segment is a decreasing concave up curve that approaches infinity as x goes to 10 and approaches 0 as x approaches infinity.\" width=\"325\" height=\"246\" \/><figcaption class=\"wp-caption-text\">Figure 3. The domain of this function is [latex]x\\ne 10.[\/latex]<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of [latex]\\dfrac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572480517\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572480517\" class=\"hidden-answer\" style=\"display: none\">\nThis can be rewritten as [latex]\\displaystyle\\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx.[\/latex] <\/p>\n<p>Use substitution. Let [latex]u={x}^{4}+3{x}^{2},[\/latex] then [latex]du=4{x}^{3}+6x.[\/latex] Alter <em>du<\/em> by factoring out the [latex]2[\/latex]. <\/p>\n<p>Thus,<\/p>\n<div id=\"fs-id1170571807881\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{}\\\\ \\hfill du& =\\hfill & (4{x}^{3}+6x)dx\\hfill \\\\ & =\\hfill & 2(2{x}^{3}+3x)dx\\hfill \\\\ \\hfill \\frac{1}{2}du& =\\hfill & (2{x}^{3}+3x)dx.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572373464\">Rewrite the integrand in [latex]u[\/latex]:<\/p>\n<div id=\"fs-id1170572331833\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx=\\frac{1}{2}\\displaystyle\\int {u}^{-1}du.[\/latex]<\/div>\n<p id=\"fs-id1170571733965\">Then we have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}\\frac{1}{2}{\\displaystyle\\int {u}^{-1}du}\\hfill & =\\frac{1}{2}\\text{ln}|u|+C\\hfill \\\\ & =\\frac{1}{2}\\text{ln}|{x}^{4}+3{x}^{2}|+C.\\hfill \\end{array}[\/latex]<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8mW2bG6HoPE?controls=0&amp;start=162&amp;end=270&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.6.2_162to270_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.6.2&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the antiderivative of the log function [latex]{\\text{log}}_{2}x.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571649954\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571649954\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571649954\">Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have<\/p>\n<div id=\"fs-id1170571649959\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int {\\text{log}}_{2}xdx=\\frac{x}{\\text{ln}2}(\\text{ln}x-1)+C.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>The example below is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.<\/p>\n<section class=\"textbox example\">\n<p>Find the definite integral of [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}\\frac{ \\sin x}{1+ \\cos x}dx.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571636148\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571636148\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571636148\">We need substitution to evaluate this problem. Let [latex]u=1+ \\cos x,,[\/latex] so [latex]du=\\text{\u2212} \\sin xdx.[\/latex] Rewrite the integral in terms of [latex]u[\/latex], changing the limits of integration as well. Thus,<\/p>\n<div id=\"fs-id1170571636203\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}u=1+ \\cos (0)=2\\hfill \\\\ u=1+ \\cos (\\frac{\\pi }{2})=1\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572510108\">Then<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int} _{0}^{\\pi \\text{\/}2}\\dfrac{ \\sin x}{1+ \\cos x}\\hfill & =\\text{\u2212}{\\displaystyle\\int }_{2}^{1}{u}^{-1}du\\hfill&nbsp; \\\\ & ={\\displaystyle\\int }_{1}^{2}{u}^{-1}du\\hfill \\\\ & ={\\text{ln}|u||}_{1}^{2}\\hfill \\\\ & =\\left[\\text{ln}2-\\text{ln}1\\right]\\hfill \\\\ & =\\text{ln}2\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288437\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288437&theme=lumen&iframe_resize_id=ohm288437&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.6.2\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":399,"module-header":"","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"5.6.2","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/409"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/409\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/399"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/409\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=409"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=409"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=409"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=409"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}