# How do you evaluate the indefinite integral of dx/(81+x^2)^2?

May 11, 2015

Hi, here the polynome is irreducible you can't do partial fraction so we will use substitution

Let's $x = 9 \tan \left(u\right)$
$\mathrm{dx} = 9 {\sec}^{2} \left(u\right) \mathrm{du}$
$u = \arctan \left(\frac{1}{9} x\right)$

So we have :

$\int \frac{1}{81 + {x}^{2}} ^ 2 \mathrm{dx} = \int \frac{1}{81 + 81 {\tan}^{2} \left(u\right)} ^ 2 \cdot {\sec}^{2} \left(u\right) \mathrm{du}$

$= \frac{1}{7291} \int \frac{1}{1 + {\tan}^{2} \left(u\right)} ^ 2 \cdot {\sec}^{2} \left(u\right) \mathrm{du}$

Don't forget $1 + {\tan}^{2} \left(u\right) = {\sec}^{2} \left(u\right)$

$\frac{1}{729} \int {\cos}^{4} \left(u\right) \cdot {\sec}^{2} \left(u\right) \mathrm{du} = \frac{1}{729} \int {\cos}^{2} \left(u\right) \mathrm{du}$

Don't forget ${\cos}^{2} \left(u\right) = \frac{1}{2} \left(1 + \cos \left(2 u\right)\right)$

$\frac{1}{1458} \int 1 + \cos \left(2 u\right) \mathrm{du} = \frac{1}{2916} \int 2 + 2 \cos \left(2 u\right) \mathrm{du}$

$\frac{1}{2916} \left[2 u + \sin \left(2 u\right)\right] + C$

$\frac{1}{2916} \left[2 \arctan \left(\frac{1}{9} x\right) + \sin \left(2 \arctan \left(\frac{1}{9} x\right)\right)\right]$

$\sin \left(\arctan \left(x\right)\right) = \frac{x}{\sqrt{{x}^{2} + 1}}$

so we have

$\frac{1}{1458} \left[\arctan \left(\frac{1}{9} x\right) + \frac{9 x}{\sqrt{{x}^{2} + 81}}\right] + C$