{"id":713,"date":"2025-06-20T17:07:06","date_gmt":"2025-06-20T17:07:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=713"},"modified":"2025-07-17T15:01:08","modified_gmt":"2025-07-17T15:01:08","slug":"partial-fractions-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/partial-fractions-learn-it-1\/","title":{"raw":"Partial Fractions: Learn It 1","rendered":"Partial Fractions: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Break down and integrate rational functions using partial fractions<\/li>\r\n \t<li>Identify and work with simple linear factors in rational functions<\/li>\r\n \t<li>Handle repeated linear factors when using partial fractions<\/li>\r\n \t<li>Work with quadratic factors in rational functions<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>What Is Partial Fraction Decomposition?<\/h2>\r\n<p class=\"whitespace-normal break-words\">Partial fraction decomposition is a technique that lets you rewrite complicated rational functions as sums of simpler fractions. Think of it as \"reverse engineering\" fractions\u2014instead of adding fractions together, you're splitting them apart.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">For example, you can rewrite [latex]\\frac{3x}{{x}^{2}-x - 2}[\/latex] as [latex]\\frac{1}{x+1}+\\frac{2}{x - 2}[\/latex].<\/section>\r\n<p class=\"whitespace-normal break-words\">Why does this matter? Because integrating simpler fractions is much easier than tackling the original complex fraction.<\/p>\r\n<p class=\"whitespace-normal break-words\">You already know how to integrate basic rational functions:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\int \\frac{du}{u}=\\text{ln}|u|+C[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\int \\frac{du}{{u}^{2}+{a}^{2}}=\\frac{1}{a}{\\tan}^{-1}\\left(\\frac{u}{a}\\right)+C[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">But what about something like [latex]\\displaystyle\\int \\frac{3x}{{x}^{2}-x - 2}dx[\/latex]? This doesn't fit our basic patterns.<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">However, if we can decompose it into simpler parts:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{3x}{{x}^{2}-x - 2}dx = \\displaystyle\\int \\left(\\frac{1}{x+1}+\\frac{2}{x - 2}\\right)dx[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Now we can integrate each piece separately:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\left(\\frac{1}{x+1}+\\frac{2}{x - 2}\\right)dx=\\text{ln}|x+1|+2\\text{ln}|x - 2|+C[\/latex]<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Check Your Work<\/strong>: You can verify that [latex]\\frac{1}{x+1}+\\frac{2}{x - 2}=\\frac{3x}{{x}^{2}-x - 2}[\/latex] by finding a common denominator.<\/section>\r\n<h3>When Can You Use This Method?<\/h3>\r\nPartial fraction decomposition works only when the degree of the numerator is <strong>less than<\/strong> the degree of the denominator.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>degree requirement for partial fraction decomposition<\/h3>\r\nFor a rational function [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex], you can use partial fractions only if [latex]\\text{deg}\\left(P\\left(x\\right)\\right)&lt;\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex].\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"whitespace-normal break-words\"><strong>What if the degree of the numerator is greater than or equal to the degree of the denominator?<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">You need to use polynomial long division first. This rewrites [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] as [latex]A\\left(x\\right)+\\frac{R\\left(x\\right)}{Q\\left(x\\right)}[\/latex], where [latex]\\text{deg}\\left(R\\left(x\\right)\\right)&lt;\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Then you can apply partial fraction decomposition to [latex]\\frac{R(x)}{Q(x)}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Polynomial Long Division<\/strong>\r\n<ol>\r\n \t<li>Set up the division problem as the numerator divided by the denominator<\/li>\r\n \t<li>Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.<\/li>\r\n \t<li>Multiply the answer by the divisor and write it below the like terms of the dividend.<\/li>\r\n \t<li>Subtract the bottom binomial from the top binomial.<\/li>\r\n \t<li>Bring down the next term of the dividend.<\/li>\r\n \t<li>Repeat steps 2\u20135 until reaching the last term of the dividend.<\/li>\r\n \t<li>If the remainder is non-zero, express as a fraction using the divisor as the denominator.<\/li>\r\n<\/ol>\r\nVisit <a href=\"https:\/\/en.wikipedia.org\/wiki\/Polynomial_long_division\" target=\"_blank\" rel=\"noopener\">this website for a review of long division of polynomials<\/a>.\r\n\r\n<\/section>The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form [latex]\\displaystyle\\int \\frac{P\\left(x\\right)}{Q\\left(x\\right)}dx[\/latex], where [latex]\\text{deg}\\left(P\\left(x\\right)\\right)\\ge \\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex].\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042044846\" data-type=\"problem\">\r\n<p id=\"fs-id1165041845429\">Evaluate [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+5}{x+1}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1165040794978\" data-type=\"solution\">\r\n<p id=\"fs-id1165041796692\">Since [latex]\\text{deg}\\left({x}^{2}+3x+5\\right)\\ge \\text{deg}\\left(x+1\\right)[\/latex], we perform long division to obtain<\/p>\r\n\r\n<div id=\"fs-id1165040774466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{2}+3x+5}{x+1}=x+2+\\frac{3}{x+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040796470\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042278063\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int \\frac{{x}^{2}+3x+5}{x+1}dx}&amp; ={\\displaystyle\\int \\left(x+2+\\frac{3}{x+1}\\right)dx}\\hfill \\\\ &amp; =\\frac{1}{2}{x}^{2}+2x+3\\text{ln}|x+1|+C.\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Break down and integrate rational functions using partial fractions<\/li>\n<li>Identify and work with simple linear factors in rational functions<\/li>\n<li>Handle repeated linear factors when using partial fractions<\/li>\n<li>Work with quadratic factors in rational functions<\/li>\n<\/ul>\n<\/section>\n<h2>What Is Partial Fraction Decomposition?<\/h2>\n<p class=\"whitespace-normal break-words\">Partial fraction decomposition is a technique that lets you rewrite complicated rational functions as sums of simpler fractions. Think of it as &#8220;reverse engineering&#8221; fractions\u2014instead of adding fractions together, you&#8217;re splitting them apart.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">For example, you can rewrite [latex]\\frac{3x}{{x}^{2}-x - 2}[\/latex] as [latex]\\frac{1}{x+1}+\\frac{2}{x - 2}[\/latex].<\/section>\n<p class=\"whitespace-normal break-words\">Why does this matter? Because integrating simpler fractions is much easier than tackling the original complex fraction.<\/p>\n<p class=\"whitespace-normal break-words\">You already know how to integrate basic rational functions:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\int \\frac{du}{u}=\\text{ln}|u|+C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\int \\frac{du}{{u}^{2}+{a}^{2}}=\\frac{1}{a}{\\tan}^{-1}\\left(\\frac{u}{a}\\right)+C[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">But what about something like [latex]\\displaystyle\\int \\frac{3x}{{x}^{2}-x - 2}dx[\/latex]? This doesn&#8217;t fit our basic patterns.<\/p>\n<p class=\"whitespace-pre-wrap break-words\">However, if we can decompose it into simpler parts:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{3x}{{x}^{2}-x - 2}dx = \\displaystyle\\int \\left(\\frac{1}{x+1}+\\frac{2}{x - 2}\\right)dx[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Now we can integrate each piece separately:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\int \\left(\\frac{1}{x+1}+\\frac{2}{x - 2}\\right)dx=\\text{ln}|x+1|+2\\text{ln}|x - 2|+C[\/latex]<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Check Your Work<\/strong>: You can verify that [latex]\\frac{1}{x+1}+\\frac{2}{x - 2}=\\frac{3x}{{x}^{2}-x - 2}[\/latex] by finding a common denominator.<\/section>\n<h3>When Can You Use This Method?<\/h3>\n<p>Partial fraction decomposition works only when the degree of the numerator is <strong>less than<\/strong> the degree of the denominator.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>degree requirement for partial fraction decomposition<\/h3>\n<p>For a rational function [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex], you can use partial fractions only if [latex]\\text{deg}\\left(P\\left(x\\right)\\right)<\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex].\n\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"whitespace-normal break-words\"><strong>What if the degree of the numerator is greater than or equal to the degree of the denominator?<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">You need to use polynomial long division first. This rewrites [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] as [latex]A\\left(x\\right)+\\frac{R\\left(x\\right)}{Q\\left(x\\right)}[\/latex], where [latex]\\text{deg}\\left(R\\left(x\\right)\\right)<\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Then you can apply partial fraction decomposition to [latex]\\frac{R(x)}{Q(x)}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Polynomial Long Division<\/strong><\/p>\n<ol>\n<li>Set up the division problem as the numerator divided by the denominator<\/li>\n<li>Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.<\/li>\n<li>Multiply the answer by the divisor and write it below the like terms of the dividend.<\/li>\n<li>Subtract the bottom binomial from the top binomial.<\/li>\n<li>Bring down the next term of the dividend.<\/li>\n<li>Repeat steps 2\u20135 until reaching the last term of the dividend.<\/li>\n<li>If the remainder is non-zero, express as a fraction using the divisor as the denominator.<\/li>\n<\/ol>\n<p>Visit <a href=\"https:\/\/en.wikipedia.org\/wiki\/Polynomial_long_division\" target=\"_blank\" rel=\"noopener\">this website for a review of long division of polynomials<\/a>.<\/p>\n<\/section>\n<p>The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form [latex]\\displaystyle\\int \\frac{P\\left(x\\right)}{Q\\left(x\\right)}dx[\/latex], where [latex]\\text{deg}\\left(P\\left(x\\right)\\right)\\ge \\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042044846\" data-type=\"problem\">\n<p id=\"fs-id1165041845429\">Evaluate [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+5}{x+1}dx[\/latex].<\/p>\n<\/div>\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<div id=\"fs-id1165040794978\" data-type=\"solution\">\n<p id=\"fs-id1165041796692\">Since [latex]\\text{deg}\\left({x}^{2}+3x+5\\right)\\ge \\text{deg}\\left(x+1\\right)[\/latex], we perform long division to obtain<\/p>\n<div id=\"fs-id1165040774466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{2}+3x+5}{x+1}=x+2+\\frac{3}{x+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040796470\">Thus,<\/p>\n<div id=\"fs-id1165042278063\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int \\frac{{x}^{2}+3x+5}{x+1}dx}& ={\\displaystyle\\int \\left(x+2+\\frac{3}{x+1}\\right)dx}\\hfill \\\\ & =\\frac{1}{2}{x}^{2}+2x+3\\text{ln}|x+1|+C.\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":20,"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 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