{"id":240827,"date":"2025-07-03T12:10:23","date_gmt":"2025-07-03T12:10:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=240827"},"modified":"2025-07-03T12:10:25","modified_gmt":"2025-07-03T12:10:25","slug":"solve-the-integral-lnsqrtx2-4-dx","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/03\/solve-the-integral-lnsqrtx2-4-dx\/","title":{"rendered":"Solve the integral: ln(sqrt(x^2 – 4)) dx"},"content":{"rendered":"\n

Solve the integral: ln(sqrt(x^2 – 4)) dx<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To solve the integral \u222bln\u2061(x2\u22124)\u2009dx\\int \\ln(\\sqrt{x^2 – 4}) \\, dx\u222bln(x2\u22124\u200b)dx, we can follow a systematic approach.<\/p>\n\n\n\n

Step 1: Simplifying the integrand<\/h3>\n\n\n\n

The first step is to simplify the expression inside the logarithm. We know that:ln\u2061(x2\u22124)=12ln\u2061(x2\u22124)\\ln(\\sqrt{x^2 – 4}) = \\frac{1}{2} \\ln(x^2 – 4)ln(x2\u22124\u200b)=21\u200bln(x2\u22124)<\/p>\n\n\n\n

So, the integral becomes:\u222bln\u2061(x2\u22124)\u2009dx=12\u222bln\u2061(x2\u22124)\u2009dx\\int \\ln(\\sqrt{x^2 – 4}) \\, dx = \\frac{1}{2} \\int \\ln(x^2 – 4) \\, dx\u222bln(x2\u22124\u200b)dx=21\u200b\u222bln(x2\u22124)dx<\/p>\n\n\n\n

Step 2: Integration by parts<\/h3>\n\n\n\n

Next, we apply the technique of integration by parts. Recall that:\u222bu\u2009dv=uv\u2212\u222bv\u2009du\\int u \\, dv = uv – \\int v \\, du\u222budv=uv\u2212\u222bvdu<\/p>\n\n\n\n

Let:<\/p>\n\n\n\n

    \n
  • u=ln\u2061(x2\u22124)u = \\ln(x^2 – 4)u=ln(x2\u22124), so that du=2xx2\u22124\u2009dxdu = \\frac{2x}{x^2 – 4} \\, dxdu=x2\u221242x\u200bdx<\/li>\n\n\n\n
  • dv=dxdv = dxdv=dx, so that v=xv = xv=x<\/li>\n<\/ul>\n\n\n\n

    Now, applying the integration by parts formula:\u222bln\u2061(x2\u22124)\u2009dx=xln\u2061(x2\u22124)\u2212\u222b2x2x2\u22124\u2009dx\\int \\ln(x^2 – 4) \\, dx = x \\ln(x^2 – 4) – \\int \\frac{2x^2}{x^2 – 4} \\, dx\u222bln(x2\u22124)dx=xln(x2\u22124)\u2212\u222bx2\u221242×2\u200bdx<\/p>\n\n\n\n

    Step 3: Simplifying the remaining integral<\/h3>\n\n\n\n

    Now, simplify the remaining integral:\u222b2x2x2\u22124\u2009dx\\int \\frac{2x^2}{x^2 – 4} \\, dx\u222bx2\u221242×2\u200bdx<\/p>\n\n\n\n

    We can use the fact that:2x2x2\u22124=2+8×2\u22124\\frac{2x^2}{x^2 – 4} = 2 + \\frac{8}{x^2 – 4}x2\u221242×2\u200b=2+x2\u221248\u200b<\/p>\n\n\n\n

    So the integral becomes:\u222b2x2x2\u22124\u2009dx=2x+8\u222b1×2\u22124\u2009dx\\int \\frac{2x^2}{x^2 – 4} \\, dx = 2x + 8 \\int \\frac{1}{x^2 – 4} \\, dx\u222bx2\u221242×2\u200bdx=2x+8\u222bx2\u221241\u200bdx<\/p>\n\n\n\n

    The integral of 1×2\u22124\\frac{1}{x^2 – 4}x2\u221241\u200b is a standard one, which is:\u222b1×2\u22124\u2009dx=14ln\u2061\u2223x\u22122x+2\u2223\\int \\frac{1}{x^2 – 4} \\, dx = \\frac{1}{4} \\ln \\left| \\frac{x – 2}{x + 2} \\right|\u222bx2\u221241\u200bdx=41\u200bln\u200bx+2x\u22122\u200b\u200b<\/p>\n\n\n\n

    So the original integral becomes:\u222bln\u2061(x2\u22124)\u2009dx=12(xln\u2061(x2\u22124)\u22122x\u22122\u22c52ln\u2061\u2223x\u22122x+2\u2223+C)\\int \\ln(\\sqrt{x^2 – 4}) \\, dx = \\frac{1}{2} \\left( x \\ln(x^2 – 4) – 2x – 2 \\cdot 2 \\ln \\left| \\frac{x – 2}{x + 2} \\right| + C \\right)\u222bln(x2\u22124\u200b)dx=21\u200b(xln(x2\u22124)\u22122x\u22122\u22c52ln\u200bx+2x\u22122\u200b\u200b+C)<\/p>\n\n\n\n

    Step 4: Final answer<\/h3>\n\n\n\n

    Simplifying the expression:\u222bln\u2061(x2\u22124)\u2009dx=x2ln\u2061(x2\u22124)\u2212x\u22122ln\u2061\u2223x\u22122x+2\u2223+C\\int \\ln(\\sqrt{x^2 – 4}) \\, dx = \\frac{x}{2} \\ln(x^2 – 4) – x – 2 \\ln \\left| \\frac{x – 2}{x + 2} \\right| + C\u222bln(x2\u22124\u200b)dx=2x\u200bln(x2\u22124)\u2212x\u22122ln\u200bx+2x\u22122\u200b\u200b+C<\/p>\n\n\n\n

    This is the final solution to the integral.<\/p>\n\n\n\n

    Conclusion<\/h3>\n\n\n\n

    By using integration by parts and simplifying the resulting terms, we arrive at the final expression for the integral of ln\u2061(x2\u22124)\\ln(\\sqrt{x^2 – 4})ln(x2\u22124\u200b). This technique relies on reducing the integrand to simpler components and utilizing standard integral formulas.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Solve the integral: ln(sqrt(x^2 – 4)) dx The Correct Answer and Explanation is: To solve the integral \u222bln\u2061(x2\u22124)\u2009dx\\int \\ln(\\sqrt{x^2 – 4}) \\, dx\u222bln(x2\u22124\u200b)dx, we can follow a systematic approach. Step 1: Simplifying the integrand The first step is to simplify the expression inside the logarithm. We know that:ln\u2061(x2\u22124)=12ln\u2061(x2\u22124)\\ln(\\sqrt{x^2 – 4}) = \\frac{1}{2} \\ln(x^2 – 4)ln(x2\u22124\u200b)=21\u200bln(x2\u22124) […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-240827","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240827","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=240827"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240827\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=240827"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=240827"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=240827"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}