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 December 9th, 2009, 09:17 AM #1 Newbie   Joined: Dec 2009 Posts: 12 Thanks: 0 Need help for an integration Can anyone help me solve this integration below? Integral of square root of (4+x^2) divided by x^5 dx. Thanks.
December 9th, 2009, 04:20 PM   #2
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Joined: Dec 2006
From: Lexington, MA

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Re: Need help for an integration

Hello, xujack88!

I made some substitutions, but I really don't want to finish it . . .

Quote:
 $\int \frac{\sqrt{4+x^2}}{x^5}\,dx$

$\text{Let: }\:x \,=\,2\,\tan\theta \qquad\Rightarrow\qquad dx \,=\,2\,\sec^2\theta\,d\theta$
[color=beige].v . [/color]$\text{and: }\:\sqrt{4+x^2} \:=\:2\,\sec\theta$

$\text{Substitute: }\;\int\frac{(2\,\sec\theta)(2\,\sec^2\theta\,d\th eta)}{32\tan^5\theta} \;=\;\frac{1}{8}\int\frac{\sec^3\theta}{\tan^5\the ta}\,d\theta$

$\text{Multiply by }\frac{\tan\theta}{\tan\theta}:\qquad\qquad \frac{1}{8}\int\frac{\tan\theta}{\tan\theta}\,\cdo t\,\frac{\sec^3\theta}{\tan^5\theta}\,d\theta \;=\;\frac{1}{8}\int \frac{\sec^3\theta\tan\theta}{\tan^6\theta}\,d\the ta$

[color=beige]. . [/color]$=\; \frac{1}{8}\int\frac{(\sec^2\theta)(\sec\theta\tan \theta\,d\theta)}{(\tan^2\theta)^3} \;=\; \frac{1}{8}\int\frac{(\sec^2\theta)}{(\sec^2\theta \,-\,1)^3}\,(\sec\theta\tan\theta\,d\theta)$

$\text{Let }u \,=\,\sec\theta \quad\Rightarrow\quad du \,=\,\sec\theta\tan\theta\,d\theta$

$\text{Subtitute: }\;\frac{1}{8}\int\frac{u^2}{(u^2\,-\,1)^3}\,du$

$\text{This can be integrated by Partial Fractions . . .$

[color=beige]. . [/color]$\frac{u^2}{(u-1)^3(u+1)^3} \;=\;\frac{A}{(u-1)}\,+\,\frac{B}{(u-1)^2}\,+\,\frac{C}{(u-1)^3} \,+\,\frac{D}{(u+1)}\,+\,\frac{E}{(u+1)^2} \,+\,\frac{F}{(u+1)^3}$

You go on ahead . . . I'll wait in the car.
[color=beige] .[/color]

 December 9th, 2009, 06:04 PM #3 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Need help for an integration There is no need for the introduction of u; the denominator (sec² ? - 1)³ is equal to (tan² ?)³. So you can turn the integrand into powers of csc ? and use a reduction formula. $\int \csc^n{\theta} \, d\theta= -\frac{\csc^{n-1}{\theta} \cos{\theta}}{n-1} \,+\, \frac{n-2}{n-1}\int \csc^{n-2}{\theta} \, d\theta \qquad \mbox{ (for }n \ne 1\mbox{)}$ It might be tricky rewriting the answer in terms of x.
 December 11th, 2009, 12:47 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 Let u = ?(4 + x²), so u² = 4 + x² and udu = xdx. ?u²/(u² - 4)³ du = ?(1/(u + 2) - 1/(u - 2) + 2/(u + 2)² + 2/(u - 2)² + 8/(u - 2)³ - 8/(u + 2)³)/128 du, which is straightforward. Your final answer (in terms of x) can be written as a logarithmic term and another term. Post your working if you want it checked.
 December 11th, 2009, 01:17 PM #5 Newbie   Joined: Dec 2009 Posts: 12 Thanks: 0 Re: Need help for an integration Thank you all guys. I learned two methods to solve this, one is by using partial fractions, the other is reduction formula. This question is actually on my final, it's kinda time-consuming to finish it.

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