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April 28th, 2016, 11:54 PM   #1
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Can any one solve this integration?

see the attachment for the question
n thanx for replying in andvance
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Last edited by jeet55; April 28th, 2016 at 11:57 PM.

 April 29th, 2016, 02:36 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,095 Thanks: 1905 $\displaystyle \int\!\frac{x^2 - 2}{x^3\sqrt{x^2 - 1}}\,dx = -\frac{\sqrt{x^2 - 1}}{x^2} + \text{C}$ Thanks from jeet55
 April 29th, 2016, 03:57 AM #3 Newbie   Joined: Aug 2015 From: india Posts: 4 Thanks: 0 Can you give a complete solution? Last edited by skipjack; April 29th, 2016 at 04:25 AM.
 April 29th, 2016, 04:28 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,095 Thanks: 1905 Differentiate my answer and simplify the result to give the original integrand. Then write down each line you wrote in reverse order, adding integral signs where appropriate. Thanks from jeet55
 April 29th, 2016, 04:37 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,551 Thanks: 2554 Math Focus: Mainly analysis and algebra I would split the numerator into $(x^2-1)-1$. Alternatively, set $u^2=x^2-1$. Thanks from jeet55 Last edited by skipjack; April 29th, 2016 at 05:09 AM.
 April 29th, 2016, 03:56 PM #6 Member   Joined: Apr 2015 From: USA Posts: 46 Thanks: 32 This one yields nicely to a trig substitution. Let $x=\sec\theta$. And $dx=\sec\theta\tan\theta\,d\theta$. Also, it helps to know that $\sqrt{\sec^2\theta-1}=\tan\theta$. Thanks from jeet55
 April 29th, 2016, 09:03 PM #7 Newbie   Joined: Aug 2015 From: india Posts: 4 Thanks: 0 thanx got it!

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