1. Integral (1+x^2)^{-2}

Hello, I'm trying to solve this integral

$\displaystyle \int \frac{dx}{(1+x^2)^2}$

and would be really grateful for any help!

2. Originally Posted by gusztav
Hello, I'm trying to solve this integral

$\displaystyle \int \frac{dx}{(1+x^2)^2}$

and would be really grateful for any help!
Did you try the sub, $\displaystyle x = \tan \theta$?

3. Why dont we generalize this problem : $\displaystyle \int \frac{dx}{(1+ x^2)^n}$

Let $\displaystyle I_{n}$ be the integral ,
$\displaystyle I_{n} = \int \frac{1 + x^2 - x^2}{(1+x^2)^n} ~dx$

$\displaystyle = \int \frac{1}{(1+x^2)^{n-1}} - \frac{x^2}{(1+x^2)^n} ~dx$

$\displaystyle = I_{n-1} - x \frac{-1}{2(n-1)(1+x^2)^{n-1}} + \frac{-1}{2(n-1)} \int \frac{dx}{(1+x^2)^{n-1}} ~~~ integration~by~parts$

$\displaystyle = I_{n-1} + \frac{x}{2(n-1)(1+x^2)^{n-1}} - \frac{1}{2(n-1)} I_{n-1}$

Therefore , $\displaystyle I_{n} = \frac{2n-3}{2n-2} I_{n-1} + \frac{x}{2(n-1)(1+x^2)^{n-1}}$

for n = 2 , $\displaystyle I_2 = \frac{1}{2} \tan^{-1}(x) + \frac{x}{2(1+x^2)^{1}}$

4. Is...

$\displaystyle \frac{1}{(1+x^{2})^{2}} = \frac{1 + x^{2} - x^{2}}{(1+x^{2})^{2}} = \frac{1}{1+x^{2}} - x\cdot \frac{x}{(1+x^{2})^{2}}$

... so that...

$\displaystyle \int \frac{dx}{(1+x^{2})^{2}} = \int \frac{dx}{1+x^{2}} + \int x\cdot \frac{x\cdot dx}{(1+x^{2})^{2}}$

The first integral is well known, the second can be solved by parts...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$