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April 28th, 2016, 10:54 PM  #1 
Newbie Joined: Aug 2015 From: india Posts: 4 Thanks: 0  Can any one solve this integration?
see the attachment for the question n thanx for replying in andvance Last edited by jeet55; April 28th, 2016 at 10:57 PM. 
April 29th, 2016, 01:36 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
$\displaystyle \int\!\frac{x^2  2}{x^3\sqrt{x^2  1}}\,dx = \frac{\sqrt{x^2  1}}{x^2} + \text{C}$

April 29th, 2016, 02:57 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 03:25 AM. 
April 29th, 2016, 03:28 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
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. 
April 29th, 2016, 03:37 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra 
I would split the numerator into $(x^21)1$. Alternatively, set $u^2=x^21$.
Last edited by skipjack; April 29th, 2016 at 04:09 AM. 
April 29th, 2016, 02:56 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 $\sqrt{\sec^2\theta1}=\tan\theta$.

April 29th, 2016, 08:03 PM  #7 
Newbie Joined: Aug 2015 From: india Posts: 4 Thanks: 0 
thanx got it!


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