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I need to evaluate the integral

$$\int \frac{1}{x\sqrt{x^2-5}} dx$$

The integral should be solved by substitution. I tried substituting $u=x^2-5$, but did not come with an answer.The correct answer is $\frac{1}{\sqrt 5}\arcsin\left(\frac{\sqrt 5}{x}\right)$.

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    $\begingroup$ You have a square root of a quadratic and you know your solution has an inverse trig function. Have you considered a trig substitution? $\endgroup$ – Mike Feb 19 '14 at 10:19
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    $\begingroup$ Use $x=\frac{\sqrt{5}}{\sin(u)}=\sqrt{5}\csc(u)$. $\endgroup$ – J.R. Feb 19 '14 at 10:27
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Hint

Use the change of variable $$x=\sqrt 5\cosh u$$

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  • $\begingroup$ Succinct and spot on! $\endgroup$ – amWhy Feb 20 '14 at 13:32
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Hint: Let's rearrange the equation as: $$ \int \frac{1}{x^2 \sqrt{1-\frac{5}{x^2}}}dx $$ This is done by pulling the $x^2$ out of the square root.

So using the substitution as provided we get $$ du = \frac{-\sqrt{5}}{x^2}dx $$ Which you should see helps you solve the integral as required. In the standard form of the trigonometric integral.

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  • $\begingroup$ how did you get du=-(5)^.5/x^2dx? what was u equal to? $\endgroup$ – Jake Feb 19 '14 at 10:42
  • $\begingroup$ I thought the substitution in the question was $u=\frac{\sqrt{5}}{x}$. After reading the OP that wasn't the sub he was using, but the one in this comment was the one I meant. $\endgroup$ – Chinny84 Feb 19 '14 at 10:49
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Like I put in my comment, the square root of a quadratic often suggests a trig substitution, which takes one of 3 forms:

$$1-\sin^2x=\cos^2x$$ $$\tan^2x+1=\sec^2x$$ $$\sec^2x-1=\tan^2x$$

In this case, you have something squared minus a constant, so the third form is what we want. The substitution $x=\sqrt5\sec t$ will reduce the square root to $\sqrt{5\tan^2t}$. If you perform the substitution correctly, you should end up integrating a constant.

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