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April 28th, 2016, 10:54 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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File Type: jpg IMG_20160429_121522_1461912360504.jpg (22.1 KB, 9 views)

Last edited by jeet55; April 28th, 2016 at 10:57 PM.
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April 29th, 2016, 01:36 AM   #2
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$\displaystyle \int\!\frac{x^2 - 2}{x^3\sqrt{x^2 - 1}}\,dx = -\frac{\sqrt{x^2 - 1}}{x^2} + \text{C}$
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April 29th, 2016, 02:57 AM   #3
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Can you give a complete solution?

Last edited by skipjack; April 29th, 2016 at 03:25 AM.
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April 29th, 2016, 03:28 AM   #4
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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.
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April 29th, 2016, 03:37 AM   #5
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I would split the numerator into $(x^2-1)-1$. Alternatively, set $u^2=x^2-1$.
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Last edited by skipjack; April 29th, 2016 at 04:09 AM.
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April 29th, 2016, 02:56 PM   #6
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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 $\sqrt{\sec^2\theta-1}=\tan\theta$.
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April 29th, 2016, 08:03 PM   #7
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thanx got it!
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