{"id":687,"date":"2025-06-20T17:03:48","date_gmt":"2025-06-20T17:03:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=687"},"modified":"2025-07-15T17:20:52","modified_gmt":"2025-07-15T17:20:52","slug":"integration-by-parts-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-by-parts-learn-it-1\/","title":{"raw":"Integration by Parts: Learn It 1","rendered":"Integration by Parts: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Recognize when to use integration by parts compared to other integration methods<\/li>\r\n \t<li>Use the integration by parts formula to solve indefinite integrals<\/li>\r\n \t<li>Apply integration by parts to evaluate definite integrals<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Integration-by-Parts<\/h2>\r\n<p class=\"whitespace-normal break-words\">You've already learned how to tackle many basic integrals, and you can handle something like [latex]\\displaystyle\\int x\\sin\\left({x}^{2}\\right)dx[\/latex] using substitution with [latex]u={x}^{2}[\/latex]. But what about [latex]\\displaystyle\\int x\\sin{x}dx[\/latex]? This seemingly simple integral requires a different approach.<\/p>\r\n<p class=\"whitespace-normal break-words\">Many students wonder if there's a product rule for integration\u2014there isn't. However, we can use a technique based on the product rule for differentiation to exchange one integral for another. This powerful method is called <strong>integration by parts<\/strong>.<\/p>\r\n<p class=\"whitespace-normal break-words\">The key insight is that when you have a product of two functions, you can often rewrite the integral in a more manageable form. Let's see how this works by deriving the integration by parts formula.<\/p>\r\n\r\n<section class=\"textbox connectIt\" aria-label=\"Connect It\">\r\n<p id=\"fs-id1165041979164\">If, [latex]h\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)[\/latex], then by using the product rule, we obtain [latex]{h}^{\\prime }\\left(x\\right)={f}^{\\prime }\\left(x\\right)g\\left(x\\right)+{g}^{\\prime }\\left(x\\right)f\\left(x\\right)[\/latex]. Although at first it may seem counterproductive, let\u2019s now integrate both sides of this equation: [latex]\\displaystyle\\int {h}^{\\prime }\\left(x\\right)dx=\\displaystyle\\int \\left(g\\left(x\\right){f}^{\\prime }\\left(x\\right)+f\\left(x\\right){g}^{\\prime }\\left(x\\right)\\right)dx[\/latex].<\/p>\r\n<p id=\"fs-id1165042262165\">This gives us:<\/p>\r\n\r\n<div id=\"fs-id1165042041848\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]h\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\int g\\left(x\\right){f}^{\\prime }\\left(x\\right)dx+\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041955261\">Now we solve for [latex]\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx:[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165042235539\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx=f\\left(x\\right)g\\left(x\\right)-\\displaystyle\\int g\\left(x\\right){f}^{\\prime }\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042135462\">By making the substitutions [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex], which in turn make [latex]du={f}^{\\prime }\\left(x\\right)dx[\/latex] and [latex]dv={g}^{\\prime }\\left(x\\right)dx[\/latex], we have the more compact form<\/p>\r\n\r\n<div id=\"fs-id1165041813626\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int udv=uv-\\displaystyle\\int vdu[\/latex].<\/div>\r\n<\/section><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<div>\r\n<h3>Integration by Parts<\/h3>\r\n<p id=\"fs-id1165042089614\">Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:<\/p>\r\n\r\n<center>[latex]\\displaystyle\\int udv=uv-\\displaystyle\\int vdu[\/latex].<\/center><\/div>\r\n<\/section><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.<\/span>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1165041832755\">Use integration by parts with [latex]u=x[\/latex] and [latex]dv=\\sin{x}dx[\/latex] to evaluate [latex]\\displaystyle\\int x\\sin{x}dx[\/latex].<\/p>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n\r\nBy choosing [latex]u=x[\/latex], we have [latex]du=1dx[\/latex]. Since [latex]dv=\\sin{x}dx[\/latex], we get [latex]v=\\displaystyle\\int \\sin{x}dx=\\text{-}\\cos{x}[\/latex]. It is handy to keep track of these values as follows:\r\n<div id=\"fs-id1165041792874\" data-type=\"solution\">\r\n<div id=\"fs-id1165041792867\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u&amp; =\\hfill &amp; x\\hfill &amp; &amp; \\hfill dv&amp; =\\hfill &amp; \\sin{x}dx\\hfill \\\\ \\hfill du&amp; =\\hfill &amp; 1dx\\hfill &amp; &amp; \\hfill v&amp; =\\hfill &amp; \\displaystyle\\int \\sin{x}dx=\\text{-}\\cos{x}. \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040681626\">Applying the integration-by-parts formula results in<\/p>\r\n\r\n<div id=\"fs-id1165041985992\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int x\\sin{x}dx}&amp; =\\left(x\\right)\\left(\\text{-}\\cos{x}\\right)-{\\displaystyle\\int \\left(\\text{-}\\cos{x}\\right)\\left(1dx\\right)}\\hfill &amp; &amp; \\text{Substitute.}\\hfill \\\\ &amp; =\\text{-}x\\cos{x}+{\\displaystyle\\int \\cos{x}dx}\\hfill &amp; &amp; \\text{Simplify.}\\hfill \\\\ &amp; =\\text{-}x\\cos{x}+\\sin{x}+C.\\hfill &amp; &amp; \\text{Use}{\\displaystyle\\int \\cos{x}dx=\\sin{x}+C.}\\end{array}[\/latex]<\/div>\r\n<div id=\"fs-id1165041893024\" data-type=\"commentary\">\r\n<div data-type=\"title\"><\/div>\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<div data-type=\"title\"><\/div>\r\n<p id=\"fs-id1165041973005\">At this point, there are probably a few items that need clarification. First of all, you may be curious about what would have happened if we had chosen [latex]u=\\sin{x}[\/latex] and [latex]dv=x[\/latex]. If we had done so, then we would have [latex]du=\\cos{x}[\/latex] and [latex]v=\\frac{1}{2}{x}^{2}[\/latex]. Thus, after applying integration by parts, we have [latex]{\\displaystyle\\int }^{\\text{ }}x\\sin{x}dx=\\frac{1}{2}{x}^{2}\\sin{x}-{\\displaystyle\\int }^{\\text{ }}\\frac{1}{2}{x}^{2}\\cos{x}dx[\/latex]. Unfortunately, with the new integral, we are in no better position than before. It is important to keep in mind that when we apply integration by parts, we may need to try several choices for [latex]u[\/latex] and [latex]dv[\/latex] before finding a choice that works.<\/p>\r\n<p id=\"fs-id1165042098353\">Second, you may wonder why, when we find [latex]v={\\displaystyle\\int }^{\\text{ }}\\sin{x}dx=\\text{-}\\cos{x}[\/latex], we do not use [latex]v=\\text{-}\\cos{x}+K[\/latex]. To see that it makes no difference, we can rework the problem using [latex]v=\\text{-}\\cos{x}+K\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165042094536\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}x\\sin{x}dx\\hfill &amp; =\\left(x\\right)\\left(\\text{-}\\cos{x}+K\\right)-{\\displaystyle\\int}\\left(\\text{-}\\cos{x}+K\\right)\\left(1dx\\right)\\hfill \\\\ \\hfill &amp; =\\text{-}x\\cos{x}+Kx+{\\displaystyle\\int}\\cos{x}dx-{\\displaystyle\\int}Kdx\\hfill \\\\ \\hfill &amp; =\\text{-}x\\cos{x}+Kx+\\sin{x}-Kx+C\\hfill \\\\ \\hfill &amp; =\\text{-}x\\cos{x}+\\sin{x}+C.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042136113\">As you can see, it makes no difference in the final solution.<\/p>\r\n<p id=\"fs-id1165041759254\">Last, we can check to make sure that our antiderivative is correct by differentiating [latex]\\text{-}x\\cos{x}+\\sin{x}+C\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165041952373\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\frac{d}{dx}\\left(\\text{-}x\\cos{x}+\\sin{x}+C\\right)\\hfill &amp; =\\left(-1\\right)\\cos{x}+\\left(\\text{-}x\\right)\\left(\\text{-}\\sin{x}\\right)+\\cos{x}\\hfill \\\\ \\hfill &amp; =x\\sin{x}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040665697\">Therefore, the antiderivative checks out.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Recognize when to use integration by parts compared to other integration methods<\/li>\n<li>Use the integration by parts formula to solve indefinite integrals<\/li>\n<li>Apply integration by parts to evaluate definite integrals<\/li>\n<\/ul>\n<\/section>\n<h2>Integration-by-Parts<\/h2>\n<p class=\"whitespace-normal break-words\">You&#8217;ve already learned how to tackle many basic integrals, and you can handle something like [latex]\\displaystyle\\int x\\sin\\left({x}^{2}\\right)dx[\/latex] using substitution with [latex]u={x}^{2}[\/latex]. But what about [latex]\\displaystyle\\int x\\sin{x}dx[\/latex]? This seemingly simple integral requires a different approach.<\/p>\n<p class=\"whitespace-normal break-words\">Many students wonder if there&#8217;s a product rule for integration\u2014there isn&#8217;t. However, we can use a technique based on the product rule for differentiation to exchange one integral for another. This powerful method is called <strong>integration by parts<\/strong>.<\/p>\n<p class=\"whitespace-normal break-words\">The key insight is that when you have a product of two functions, you can often rewrite the integral in a more manageable form. Let&#8217;s see how this works by deriving the integration by parts formula.<\/p>\n<section class=\"textbox connectIt\" aria-label=\"Connect It\">\n<p id=\"fs-id1165041979164\">If, [latex]h\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)[\/latex], then by using the product rule, we obtain [latex]{h}^{\\prime }\\left(x\\right)={f}^{\\prime }\\left(x\\right)g\\left(x\\right)+{g}^{\\prime }\\left(x\\right)f\\left(x\\right)[\/latex]. Although at first it may seem counterproductive, let\u2019s now integrate both sides of this equation: [latex]\\displaystyle\\int {h}^{\\prime }\\left(x\\right)dx=\\displaystyle\\int \\left(g\\left(x\\right){f}^{\\prime }\\left(x\\right)+f\\left(x\\right){g}^{\\prime }\\left(x\\right)\\right)dx[\/latex].<\/p>\n<p id=\"fs-id1165042262165\">This gives us:<\/p>\n<div id=\"fs-id1165042041848\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]h\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\int g\\left(x\\right){f}^{\\prime }\\left(x\\right)dx+\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041955261\">Now we solve for [latex]\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx:[\/latex]<\/p>\n<div id=\"fs-id1165042235539\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int f\\left(x\\right){g}^{\\prime }\\left(x\\right)dx=f\\left(x\\right)g\\left(x\\right)-\\displaystyle\\int g\\left(x\\right){f}^{\\prime }\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042135462\">By making the substitutions [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex], which in turn make [latex]du={f}^{\\prime }\\left(x\\right)dx[\/latex] and [latex]dv={g}^{\\prime }\\left(x\\right)dx[\/latex], we have the more compact form<\/p>\n<div id=\"fs-id1165041813626\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int udv=uv-\\displaystyle\\int vdu[\/latex].<\/div>\n<\/section>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<div>\n<h3>Integration by Parts<\/h3>\n<p id=\"fs-id1165042089614\">Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int udv=uv-\\displaystyle\\int vdu[\/latex].<\/div>\n<\/div>\n<\/section>\n<p><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.<\/span><\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1165041832755\">Use integration by parts with [latex]u=x[\/latex] and [latex]dv=\\sin{x}dx[\/latex] to evaluate [latex]\\displaystyle\\int x\\sin{x}dx[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<p>By choosing [latex]u=x[\/latex], we have [latex]du=1dx[\/latex]. Since [latex]dv=\\sin{x}dx[\/latex], we get [latex]v=\\displaystyle\\int \\sin{x}dx=\\text{-}\\cos{x}[\/latex]. It is handy to keep track of these values as follows:<\/p>\n<div id=\"fs-id1165041792874\" data-type=\"solution\">\n<div id=\"fs-id1165041792867\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccccc}\\hfill u& =\\hfill & x\\hfill & & \\hfill dv& =\\hfill & \\sin{x}dx\\hfill \\\\ \\hfill du& =\\hfill & 1dx\\hfill & & \\hfill v& =\\hfill & \\displaystyle\\int \\sin{x}dx=\\text{-}\\cos{x}. \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040681626\">Applying the integration-by-parts formula results in<\/p>\n<div id=\"fs-id1165041985992\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int x\\sin{x}dx}& =\\left(x\\right)\\left(\\text{-}\\cos{x}\\right)-{\\displaystyle\\int \\left(\\text{-}\\cos{x}\\right)\\left(1dx\\right)}\\hfill & & \\text{Substitute.}\\hfill \\\\ & =\\text{-}x\\cos{x}+{\\displaystyle\\int \\cos{x}dx}\\hfill & & \\text{Simplify.}\\hfill \\\\ & =\\text{-}x\\cos{x}+\\sin{x}+C.\\hfill & & \\text{Use}{\\displaystyle\\int \\cos{x}dx=\\sin{x}+C.}\\end{array}[\/latex]<\/div>\n<div id=\"fs-id1165041893024\" data-type=\"commentary\">\n<div data-type=\"title\"><\/div>\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<div data-type=\"title\"><\/div>\n<p id=\"fs-id1165041973005\">At this point, there are probably a few items that need clarification. First of all, you may be curious about what would have happened if we had chosen [latex]u=\\sin{x}[\/latex] and [latex]dv=x[\/latex]. If we had done so, then we would have [latex]du=\\cos{x}[\/latex] and [latex]v=\\frac{1}{2}{x}^{2}[\/latex]. Thus, after applying integration by parts, we have [latex]{\\displaystyle\\int }^{\\text{ }}x\\sin{x}dx=\\frac{1}{2}{x}^{2}\\sin{x}-{\\displaystyle\\int }^{\\text{ }}\\frac{1}{2}{x}^{2}\\cos{x}dx[\/latex]. Unfortunately, with the new integral, we are in no better position than before. It is important to keep in mind that when we apply integration by parts, we may need to try several choices for [latex]u[\/latex] and [latex]dv[\/latex] before finding a choice that works.<\/p>\n<p id=\"fs-id1165042098353\">Second, you may wonder why, when we find [latex]v={\\displaystyle\\int }^{\\text{ }}\\sin{x}dx=\\text{-}\\cos{x}[\/latex], we do not use [latex]v=\\text{-}\\cos{x}+K[\/latex]. To see that it makes no difference, we can rework the problem using [latex]v=\\text{-}\\cos{x}+K\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1165042094536\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{\\displaystyle\\int}x\\sin{x}dx\\hfill & =\\left(x\\right)\\left(\\text{-}\\cos{x}+K\\right)-{\\displaystyle\\int}\\left(\\text{-}\\cos{x}+K\\right)\\left(1dx\\right)\\hfill \\\\ \\hfill & =\\text{-}x\\cos{x}+Kx+{\\displaystyle\\int}\\cos{x}dx-{\\displaystyle\\int}Kdx\\hfill \\\\ \\hfill & =\\text{-}x\\cos{x}+Kx+\\sin{x}-Kx+C\\hfill \\\\ \\hfill & =\\text{-}x\\cos{x}+\\sin{x}+C.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042136113\">As you can see, it makes no difference in the final solution.<\/p>\n<p id=\"fs-id1165041759254\">Last, we can check to make sure that our antiderivative is correct by differentiating [latex]\\text{-}x\\cos{x}+\\sin{x}+C\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1165041952373\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\frac{d}{dx}\\left(\\text{-}x\\cos{x}+\\sin{x}+C\\right)\\hfill & =\\left(-1\\right)\\cos{x}+\\left(\\text{-}x\\right)\\left(\\text{-}\\sin{x}\\right)+\\cos{x}\\hfill \\\\ \\hfill & =x\\sin{x}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040665697\">Therefore, the antiderivative checks out.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":7,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":541,"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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/687"}],"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\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/687\/revisions"}],"predecessor-version":[{"id":1182,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/687\/revisions\/1182"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/541"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/687\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=687"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=687"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=687"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=687"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}