# Thread: How to solve this integral?

1. ## How to solve this integral?

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:

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

It should be easy to solve now.

3. Hello, trigoon!

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

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

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

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

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

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