Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Calculus and Beyond Homework Help
Stuck on this integral (using partial fraction decomposition)
Reply to thread
Message
[QUOTE="QaH, post: 5891870, member: 636371"] [h2]Homework Statement [/h2] [tex]\int\frac{x^2}{\sqrt{x^2+4}}dx[/tex] [h2]Homework Equations[/h2] n/a [h2]The Attempt at a Solution[/h2] Letting [tex]x=2tan\theta[/tex] and [tex]dx=2sec^2\theta d\theta[/tex] [tex]\int\frac{x^2}{\sqrt{x^2+4}}dx=\int\frac{4tan^2\theta}{\sqrt{4+4tan^2\theta}}2sec^2\theta d\theta=\int\frac{8tan^2\theta sec^2\theta}{\sqrt{4(1+tan^2\theta)}}d\theta=4\int\frac{tan^2\theta sec^2\theta}{sec\theta}d\theta=[/tex][tex]4\int tan^2\theta sec\theta d\theta=4\int(sec^2\theta -1)sec\theta d\theta=4\int (sec^3\theta-sec\theta)d\theta=[/tex][tex]4\int sec^3\theta d\theta-4\int sec\theta d\theta=-4\int\frac{cos\theta}{cos^2\theta}d\theta+4\int\frac{cos\theta}{cos^4\theta}d\theta= [/tex][tex]-4\int\frac{cos\theta}{1-sin^2\theta}d\theta+4\int\frac{cos\theta}{(cos^2\theta)^2}d\theta=-4\int\frac{cos\theta}{1-sin^2\theta}d\theta+4\int\frac{cos\theta}{(1-sin^2\theta)^2}d\theta[/tex] now letting [tex]u=sin\theta[/tex]and[tex]du=cos\theta d\theta[/tex][tex]-4\int\frac{1}{1-u^2}du+4\int\frac{1}{(1-u^2)^2}du=-4\int\frac{1}{(1-u)(1+u)}du+4\int\frac{1}{(1-u^2)^2}du[/tex] using partial fraction decomposition I get [tex]\frac{1}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}[/tex] multipying both sides by [tex]1-u^2[/tex][tex]1=A(1+u)+B(1-u)[/tex] Letting u=1 we get that [tex]A=\frac{1}{2}[/tex] now letting u=-1 we get that [tex]B=\frac{1}{2}[/tex] Then [tex]\frac{1}{(1-u)(1+u)}=\frac{1}{2(1-u)}+\frac{1}{2(1+u)}[/tex] and [tex]-4\int\frac{1}{(1-u)(1+u)}du[/tex] becomes [tex]-4\int\frac{1}{2}(\frac{1}{1-u}+\frac{1}{1+u})du=-2(\ln\mid1+u\mid-\ln\mid1-u\mid+c[/tex]note that [tex]u=sin\theta=\frac{x}{\sqrt{x^2+4}}[/tex][tex]-2(\ln\mid1+\frac{x}{\sqrt{x^2+4}}\mid-\ln\mid1-\frac{x}{\sqrt{x^2+4}}\mid)+c=-2\ln\mid1+\frac{x}{\sqrt{x^2+4}}\mid+2\ln\mid1-\frac{x}{\sqrt{x^2+4}}\mid+c=\ln\mid(1-\frac{x}{\sqrt{x^2+4}})^2\mid-\ln\mid(1+\frac{x}{\sqrt{x^2+4}})^{2}\mid+c=\ln\mid\frac{(1-\frac{x}{\sqrt{x^2+4}})^2}{(1+\frac{x}{\sqrt{x^2+4}})^2}\mid+c[/tex] and our original integral becomes [tex]\ln\mid\frac{(1-\frac{x}{\sqrt{x^2+4}})^2}{(1+\frac{x}{\sqrt{x^2+4}})^2}\mid+4\int\frac{1}{(1-u^2)^2}du[/tex] now this is where I am stuck because I can't seem to figue out partial fraction decomposition on [tex]\frac{1}{(1-u^2)^2}=\frac{A}{(1-u^2)}+\frac{B}{(1-u^2)^2}[/tex] multiplying through I get that [tex]1=A(1-u^2)+B[/tex] Letting u=1 B=1 then[tex]1=A(1-u^2)+1[/tex] then for any u A=0 and I am left with [tex]\frac{1}{(1-u^2)^2}=0+\frac{1}{(1-u^2)^2}[/tex] and I still can't integrate. Please don't just say I am not using the correct partial fraction method because I have searched and searched but can't find a method on this type of fraction. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Stuck on this integral (using partial fraction decomposition)
Back
Top