Integral of 1/(1+x^2)^3

Evaluate $\int \frac{1}{(x^2 + 1)^3} \; dx $

Answer:

Substitute $x = \tan \theta$ and $dx = \sec^2 \theta \; d\theta $. So,

$\int \frac{1}{(x^2 + 1)^3} \; dx = \left( \int \frac{\sec^2 \theta \; d\theta}{\sec^4 \theta}\right) $

There is a typo in the above line caught by +John Baez  The exponent in bottom is 6.



$= \int \frac{\sec^2 \theta \; d\theta}{\sec^6 \theta} $

$= \int \cos^4 d\theta $

$= \int (\cos^2 \theta)^2 \; d\theta $

$= \int \left( \frac{1 + \cos 2\theta}{2} \right)^2 \; d\theta $

$= \frac{1}{4} \int (1 + 2\cos 2\theta + \cos^2 2\theta) \; d\theta $

$= \frac{1}{4}\theta + \frac{1}{4} \sin 2\theta + \frac{1}{4} \int \cos^2 2\theta \; d\theta $

$= \frac{1}{4}\theta + \frac{1}{4} \sin 2\theta + \frac{1}{4} \int \left( \frac{1 + \cos 4\theta}{2} \right) \; d\theta $

$= \frac{1}{4}\theta + \frac{1}{4} \sin 2\theta + \frac{1}{8} \left(1 + \cos 4\theta \right) \; d\theta $

$= \frac{1}{4}\theta + \frac{1}{4} \sin 2\theta + \frac{1}{8} \left(\theta + \frac{1}{4} \sin 4\theta \right) + C $

$= \frac{3}{8}\theta + \frac{1}{4} \sin 2\theta + \frac{1}{32} \sin 4\theta + C $

$\sec^2 \theta = 1 + x^2 \;,\; \cos^2 \theta = \frac{1}{1 + x^2} \;,\; \sin^2 \theta = \frac{2x}{1 + x^2} $

and $\sin^2 2\theta = 4\cos^2 \theta \sin^2 \theta = \frac{4x^2}{(1 + x^2)^2} $

and $\cos 2\theta = \frac{1-x^2}{1 + x^2} \;,\; \sin 2\theta = \frac{2x}{1+x^2} $

and $\sin 4\theta = 2sin 2\theta \cos 2\theta = \frac{4x(1-x^2)}{(1 + x^2)^2} $

So,

$\int \frac{1}{(x^2 + 1)^3} \; dx $

$= \frac{3}{8}\theta + \frac{1}{4} \sin 2\theta + \frac{1}{32} \sin 4\theta + C $

$= \frac{3}{8} \tan^{-1}x + \frac{1}{4}\frac{2x}{1+x^2} + \frac{1}{32} \frac{4x(1-x^2)}{(1 + x^2)^2} + C $

$= \frac{3}{8} \tan^{-1}x + \frac{1}{2}\frac{x}{1+x^2} + \frac{1}{8} \frac{x(1-x^2)}{(1 + x^2)^2} + C$

$= \frac{3}{8} \tan^{-1}x + \frac{x}{8(1 + x^2)^2} ( 4(1+x^2) +(1-x^2)) +C $

$= \frac{3}{8} \tan^{-1}x + \frac{x(5+3x^2)}{8(1 + x^2)^2} + C $