How to solve this integral?

**trigoon** · December 4th 2008, 10:21 AM

Firstly thank you everyone for all your help on my other questions, I'm beginning to get how to get to the answers but still have a bunch of problems so I'll try and post them if I can't get the answer.

For this question I'm not even sure what method to use:

**Chop Suey** · December 4th 2008, 10:49 AM

Make the sub $u = \frac{1}{x}$ which yields:
$- \int \frac{u}{\sqrt{25u^2+1}}~du$

It should be easy to solve now.

**Soroban** · December 4th 2008, 10:52 AM

Hello, trigoon!

$\int\frac{dx}{x^2\sqrt{x^2+25}}$

Let: $x \:=\:5\tan\theta \quad\Rightarrow\quad dx \:=\:5\sec^2\!\theta\,d\theta$ . . . and: . $\sqrt{x^2+25} \:=\:5\sec\theta$

Substitute: . $\int\frac{5\sec^2\!\theta\,d\theta}{(25\tan^2\!\th eta\cdot5\sec\theta} \;=\;\frac{1}{25}\int\frac{\sec\theta}{\tan^2\!\th eta}\,d\theta$ . $= \;\frac{1}{25}\int\frac{\frac{1}{\cos\theta}}{\fra c{\sin^2\!\theta}{\cos^2\!\theta}} \;=\;\frac{1}{25}\int\frac{\cos\theta}{\sin^2\!\th eta}\,d\theta$

. . $= \;\frac{1}{25}\int\frac{1}{\sin\theta}\cdot\frac{\ cos\theta}{\sin\theta}\,d\theta \;=\;\frac{1}{25}\int\csc\theta\cot\theta\,d\theta \;=\;-\frac{1}{25}\csc\theta + C$

Back-substitute: . $\tan\theta \:=\:\frac{x}{5} \quad\Rightarrow\quad \csc\theta \:=\:\frac{\sqrt{x^2+25}}{x}$

. . Answer: . $-\frac{\sqrt{x^2+25}}{25x} + C$

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