{"id":1965,"date":"2024-01-02T19:24:26","date_gmt":"2024-01-02T17:24:26","guid":{"rendered":"https:\/\/matematicastro.es\/?p=1965"},"modified":"2025-02-10T14:48:36","modified_gmt":"2025-02-10T12:48:36","slug":"una-integral-racional","status":"publish","type":"post","link":"https:\/\/matematicastro.es\/?p=1965","title":{"rendered":"Una integral racional"},"content":{"rendered":"\n

Conviene aprender hasta del enemigo<\/em><\/p>\n\n\n\n

Ovidio<\/a><\/em><\/p>\n\n\n\n

\n\n\n\n

Vamos a calcular una primitiva de la funci\u00f3n \\(f(x)=\\dfrac{1}{x^2-a^2}\\) donde \\(a\\) es un n\u00famero real cualquiera distinto de cero. Es decir, se trata de calcular la integral indefinida \\(\\displaystyle\\int{\\frac{1}{x^2-a^2}dx}\\). Para ello vamos a descomponer en dos fracciones simples la fracci\u00f3n \\(\\dfrac{1}{x^2-a^2}\\). Como \\(x^2-a^2=(x+a)(x-a)\\), tenemos:<\/p>\n\n\n\n

\\[\\frac{1}{x^2-a^2}=\\frac{E}{x+a}+\\frac{F}{x-a}=\\frac{E(x-a)+F(x+a)}{(x+a)(x-a)}=\\]<\/p>\n\n\n\n

\\[=\\frac{Ex-Ea+Fx+Fa}{x^2-a^2}=\\frac{(E+F)x-Ea+Fa}{x^2-a^2}\\]<\/p>\n\n\n\n

De aqu\u00ed se deduce, igualando las fracciones algebraicas primera y \u00faltima, que<\/p>\n\n\n\n

\\[\\begin{cases}E+F=0\\\\-Ea+Fa=1\\end{cases}\\Rightarrow\\begin{cases}E=-F\\\\2Fa=1\\end{cases}\\Rightarrow\\begin{cases}E=-\\frac{1}{2a}\\\\F=\\frac{1}{2a}\\end{cases}\\]<\/p>\n\n\n\n

Es decir:<\/p>\n\n\n\n

\\[\\frac{1}{x^2-a^2}=\\frac{-\\frac{1}{2a}}{x+a}+\\frac{\\frac{1}{2a}}{x-a}\\]<\/p>\n\n\n\n

Por tanto:<\/p>\n\n\n\n

\\[\\int{\\frac{1}{x^2-a^2}dx}=\\int{\\frac{-\\frac{1}{2a}}{x+a}dx}+\\int{\\frac{\\frac{1}{2a}}{x-a}dx}=\\]<\/p>\n\n\n\n

\\[=-\\frac{1}{2a}\\int{\\frac{1}{x+a}dx}+\\frac{1}{2a}\\int{\\frac{1}{x-a}dx}=-\\frac{1}{2a}\\ln|x+a|+\\frac{1}{2a}\\ln|x-a|+C=\\]<\/p>\n\n\n\n

\\[=\\frac{1}{2a}(\\ln|x-a|-\\ln|x+a|)+C=\\frac{1}{2a}\\ln\\left|\\frac{x-a}{x+a}\\right|+C\\]<\/p>\n\n\n\n

O sea:<\/p>\n\n\n\n

\\[\\int{\\frac{1}{x^2-a^2}dx}=\\frac{1}{2a}\\ln\\left|\\frac{x-a}{x+a}\\right|+C\\qquad(1)\\]<\/p>\n\n\n\n

Ejemplo<\/h3>\n\n\n\n

Calcular \\(\\displaystyle\\int{\\frac{2}{4x^2-1}dx}\\)<\/p>\n\n\n\n

Soluci\u00f3n.<\/strong><\/p>\n\n\n\n

\\[\\int{\\frac{2}{4x^2-1}dx}=2\\int{\\frac{1}{4x^2-1}dx}=2\\int{\\frac{\\frac{1}{4}}{x^2-\\frac{1}{4}}dx}=\\]<\/p>\n\n\n\n

\\[=2\\cdot\\frac{1}{4}\\int{\\frac{1}{x^2-\\left(\\frac{1}{2}\\right)^2}dx}=\\frac{1}{2}\\int{\\frac{1}{x^2-\\left(\\frac{1}{2}\\right)^2}dx}\\]<\/p>\n\n\n\n

La integral \\(\\displaystyle\\int{\\frac{1}{x^2-\\left(\\frac{1}{2}\\right)^2}dx}\\) es del tipo \\(\\displaystyle\\int{\\frac{1}{x^2-a^2}dx}\\) con \\(a=\\dfrac{1}{2}\\). Por tanto, usando la f\u00f3rmula \\((1)\\):<\/p>\n\n\n\n

\\[\\int{\\frac{1}{x^2-\\left(\\frac{1}{2}\\right)^2}dx}=\\frac{1}{2\\cdot\\frac{1}{2}}\\ln\\left|\\frac{x-\\frac{1}{2}}{x+\\frac{1}{2}}\\right|+C=\\ln\\left|\\frac{2x-1}{2x+1}\\right|+C\\]<\/p>\n\n\n\n

Entonces:<\/p>\n\n\n\n

\\[\\int{\\frac{2}{4x^2-1}dx}=\\frac{1}{2}\\int{\\frac{1}{x^2-\\left(\\frac{1}{2}\\right)^2}dx}=\\frac{1}{2}\\ln\\left|\\frac{2x-1}{2x+1}\\right|+C\\]<\/p>\n","protected":false},"excerpt":{"rendered":"

Una integral racional que se resuelve descomponiendo la fracci\u00f3n algebraica original en fracciones simples. El resultado es un logaritmo.<\/p>\n","protected":false},"author":1,"featured_media":1967,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5,47,4],"tags":[48,49,52,51,53,50],"class_list":{"0":"post-1965","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","6":"hentry","7":"category-analisis-matematico","8":"category-analisis-matematico-2","9":"category-bachillerato","10":"tag-integral","11":"tag-integral-indefinida","12":"tag-integral-racional","13":"tag-integrales","14":"tag-primitiva","15":"tag-primitivas","17":"post-with-thumbnail","18":"post-with-thumbnail-large"},"_links":{"self":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts\/1965","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1965"}],"version-history":[{"count":3,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts\/1965\/revisions"}],"predecessor-version":[{"id":2813,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts\/1965\/revisions\/2813"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/media\/1967"}],"wp:attachment":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1965"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1965"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1965"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}