{“appState”:{“pageLoadApiCallsStatus”:true},”articleState”:{“article”:{“headers”:{“creationTime”:”2016-03-26T21:19:03+00:00″,”modifiedTime”:”2016-03-26T21:19:03+00:00″,”timestamp”:”2022-09-14T18:09:58+00:00″},”data”:{“breadcrumbs”:[{“name”:”Academics & The Arts”,”_links”:{“self”:”https://dummies-api.dummies.com/v2/categories/33662″},”slug”:”academics-the-arts”,”categoryId”:33662},{“name”:”Math”,”_links”:{“self”:”https://dummies-api.dummies.com/v2/categories/33720″},”slug”:”math”,”categoryId”:33720},{“name”:”Calculus”,”_links”:{“self”:”https://dummies-api.dummies.com/v2/categories/33723″},”slug”:”calculus”,”categoryId”:33723}],”title”:”How to Integrate by Using Partial Fractions when the Denominator Contains Only Linear Factors”,”strippedTitle”:”how to integrate by using partial fractions when the denominator contains only linear factors”,”slug”:”how-to-integrate-by-using-partial-fractions-when-the-denominator-contains-only-linear-factors”,”canonicalUrl”:””,”seo”:{“metaDescription”:”You can use the partial fractions method to integrate rational functions (Recall that a rational function is one polynomial divided by another.) The basic idea “,”noIndex”:0,”noFollow”:0},”content”:”<p>You can use the partial fractions method to integrate rational functions (Recall that a rational function is one polynomial divided by another.) The basic idea behind the partial fraction approach is “unadding” a fraction: </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204814.image0.png\” width=\”520\” height=\”104\” alt=\”image0.png\”/>\n<p>Before using the partial fractions technique, you have to check that your integrand is a “proper” fraction — that’s one where the degree of the numerator is less than the degree of the denominator. If the integrand is “improper,” like </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204815.image1.png\” width=\”119\” height=\”49\” alt=\”image1.png\”/>\n<p>you have to first do long polynomial division to transform the improper fraction into a sum of a polynomial (which sometimes will just be a number) and a proper fraction. Here’s the division for this improper fraction (without explanation). Basically, it works just like regular long division.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204816.image2.png\” width=\”199\” height=\”104\” alt=\”image2.png\”/>\n<p>With regular division, if you divide 4 into 23, you get a quotient of 5 and a remainder of 3, which tells you that </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204817.image3.png\” width=\”155\” height=\”41\” alt=\”image3.png\”/>\n<p>The result of the above polynomial division tells you the same thing. </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204818.image4.png\” width=\”493\” height=\”144\” alt=\”image4.png\”/>\n<p>The first integral is just 2<i>x</i>. You would then do the second integral with the partial fractions method.</p>\n<p>Here’s how the method works, but let’s tackle a less complicated integral than the one immediately above; this will make the technique easier to follow. </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204819.image5.png\” width=\”164\” height=\”45\” alt=\”image5.png\”/>\n<ol class=\”level-one\”>\n <li><p class=\”first-para\”>Factor the denominator.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204820.image6.png\” width=\”167\” height=\”44\” alt=\”image6.png\”/>\n </li>\n <li><p class=\”first-para\”>Break up the fraction on the right into a sum of fractions, where each factor of the denominator in Step 1 becomes the denominator of a separate fraction. Then put unknowns in the numerator of each fraction.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204821.image7.png\” width=\”203\” height=\”44\” alt=\”image7.png\”/>\n </li>\n <li><p class=\”first-para\”>Multiply both sides of this equation by the denominator of the left side. </p>\n<p class=\”child-para\”>This is algebra I, so you can’t possibly want to see the steps. Right?</p>\n<p class=\”child-para\”>5 = <i>A</i>(<i>x</i> + 3) + <i>B</i>(<i>x</i> – 2)</p>\n </li>\n <li><p class=\”first-para\”>Take the roots of the linear factors and plug them — one at a time — into <i>x</i> in the equation from Step 3, and solve for the unknowns.</p>\n<p class=\”child-para\”>If <i>x</i> = 2</p>\n<p class=\”child-para\”>5 = <i>A</i>(2 + 3) + <i>B</i>(2 – 2)</p>\n<p class=\”child-para\”>5 = 5<i>A</i></p>\n<p class=\”child-para\”><i>A</i> = 1</p>\n<p class=\”child-para\”>Or if <i>x</i> = –3</p>\n<p class=\”child-para\”>5 = <i>A</i>(–3 + 3) +<i> B</i>(–3 – 2)</p>\n<p class=\”child-para\”>5 = –5<i>B</i></p>\n<p class=\”child-para\”><i>B</i> = –1</p>\n </li>\n <li><p class=\”first-para\”>Plug these results into the <i>A</i> and <i>B</i> in the equation from Step 2.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204822.image8.png\” width=\”203\” height=\”44\” alt=\”image8.png\”/>\n </li>\n <li><p class=\”first-para\”>Split up the original integral into the partial fractions from Step 5 and you’re home free.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204823.image9.png\” width=\”352\” height=\”117\” alt=\”image9.png\”/>\n </li>\n</ol>”,”description”:”<p>You can use the partial fractions method to integrate rational functions (Recall that a rational function is one polynomial divided by another.) The basic idea behind the partial fraction approach is “unadding” a fraction: </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204814.image0.png\” width=\”520\” height=\”104\” alt=\”image0.png\”/>\n<p>Before using the partial fractions technique, you have to check that your integrand is a “proper” fraction — that’s one where the degree of the numerator is less than the degree of the denominator. If the integrand is “improper,” like </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204815.image1.png\” width=\”119\” height=\”49\” alt=\”image1.png\”/>\n<p>you have to first do long polynomial division to transform the improper fraction into a sum of a polynomial (which sometimes will just be a number) and a proper fraction. Here’s the division for this improper fraction (without explanation). Basically, it works just like regular long division.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204816.image2.png\” width=\”199\” height=\”104\” alt=\”image2.png\”/>\n<p>With regular division, if you divide 4 into 23, you get a quotient of 5 and a remainder of 3, which tells you that </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204817.image3.png\” width=\”155\” height=\”41\” alt=\”image3.png\”/>\n<p>The result of the above polynomial division tells you the same thing. </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204818.image4.png\” width=\”493\” height=\”144\” alt=\”image4.png\”/>\n<p>The first integral is just 2<i>x</i>. You would then do the second integral with the partial fractions method.</p>\n<p>Here’s how the method works, but let’s tackle a less complicated integral than the one immediately above; this will make the technique easier to follow. </p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204819.image5.png\” width=\”164\” height=\”45\” alt=\”image5.png\”/>\n<ol class=\”level-one\”>\n <li><p class=\”first-para\”>Factor the denominator.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204820.image6.png\” width=\”167\” height=\”44\” alt=\”image6.png\”/>\n </li>\n <li><p class=\”first-para\”>Break up the fraction on the right into a sum of fractions, where each factor of the denominator in Step 1 becomes the denominator of a separate fraction. Then put unknowns in the numerator of each fraction.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204821.image7.png\” width=\”203\” height=\”44\” alt=\”image7.png\”/>\n </li>\n <li><p class=\”first-para\”>Multiply both sides of this equation by the denominator of the left side. </p>\n<p class=\”child-para\”>This is algebra I, so you can’t possibly want to see the steps. Right?</p>\n<p class=\”child-para\”>5 = <i>A</i>(<i>x</i> + 3) + <i>B</i>(<i>x</i> – 2)</p>\n </li>\n <li><p class=\”first-para\”>Take the roots of the linear factors and plug them — one at a time — into <i>x</i> in the equation from Step 3, and solve for the unknowns.</p>\n<p class=\”child-para\”>If <i>x</i> = 2</p>\n<p class=\”child-para\”>5 = <i>A</i>(2 + 3) + <i>B</i>(2 – 2)</p>\n<p class=\”child-para\”>5 = 5<i>A</i></p>\n<p class=\”child-para\”><i>A</i> = 1</p>\n<p class=\”child-para\”>Or if <i>x</i> = –3</p>\n<p class=\”child-para\”>5 = <i>A</i>(–3 + 3) +<i> B</i>(–3 – 2)</p>\n<p class=\”child-para\”>5 = –5<i>B</i></p>\n<p class=\”child-para\”><i>B</i> = –1</p>\n </li>\n <li><p class=\”first-para\”>Plug these results into the <i>A</i> and <i>B</i> in the equation from Step 2.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204822.image8.png\” width=\”203\” height=\”44\” alt=\”image8.png\”/>\n </li>\n <li><p class=\”first-para\”>Split up the original integral into the partial fractions from Step 5 and you’re home free.</p>\n<img src=\”https://www.dummies.com/wp-content/uploads/204823.image9.png\” width=\”352\” height=\”117\” alt=\”image9.png\”/>\n </li>\n</ol>”,”blurb”:””,”authors”:[],”primaryCategoryTaxonomy”:{“categoryId”:33723,”title”:”Calculus”,”slug”:”calculus”,”_links”:{“self”:”https://dummies-api.dummies.com/v2/categories/33723″}},”secondaryCategoryTaxonomy”:{“categoryId”:0,”title”:null,”slug”:null,”_links”:null},”tertiaryCategoryTaxonomy”:{“categoryId”:0,”title”:null,”slug”:null,”_links”:null},”trendingArticles”:null,”inThisArticle”:[],”relatedArticles”:{“fromBook”:[],”fromCategory”:[{“articleId”:256336,”title”:”Solve a Difficult Limit Problem Using the Sandwich Method”,”slug”:”solve-a-difficult-limit-problem-using-the-sandwich-method”,”categoryList”:[“academics-the-arts”,”math”,”calculus”],”_links”:{“self”:”https://dummies-api.dummies.com/v2/articles/256336″}},{“articleId”:255765,”title”:”Solve Limit Problems on a Calculator Using Graphing 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Equations For Dummies”,”testBankPinActivationLink”:””,”bookOutOfPrint”:false,”authorsInfo”:”\n <p><p><b><b data-author-id=\”8967\”>Steven Holzner</b>, PhD,</b> taught physics at Cornell University for more than 10 years.</p>”,”authors”:[{“authorId”:8967,”name”:”Steven Holzner”,”slug”:”steven-holzner”,”description”:” <p><b>Steven Holzner, PhD,</b> taught physics at Cornell University for more than 10 years. “,”hasArticle”:false,”_links”:{“self”:”https://dummies-api.dummies.com/v2/authors/8967″}}],”_links”:{“self”:”https://dummies-api.dummies.com/v2/books/282153″}},”collections”:[],”articleAds”:{“footerAd”:”<div class=\”du-ad-region row\” id=\”article_page_adhesion_ad\”><div class=\”du-ad-unit col-md-12\” data-slot-id=\”article_page_adhesion_ad\” data-refreshed=\”false\” \r\n data-target = \”[{"key":"cat","values":["academics-the-arts","math","calculus"]},{"key":"isbn","values":[null]}]\” id=\”du-slot-632218f61a1f2\”></div></div>”,”rightAd”:”<div class=\”du-ad-region row\” 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