Question

Integrate 1/(1+x^4)^1/4 w. r. t x

Answer 1

We have to evaluate the integral $\int \frac{dx}{(1+x^4)^{1/4}}$.

First make the substitution $x^2=\tan \theta$.

$\sin \theta =\frac{x^2}{\sqrt{1+x^4}}$

Then,

$2xdx=\sec^2 \theta d\theta\\ dx=\frac{\sec^2 \theta d\theta}{2\sqrt{\tan \theta}}$

Under this substitution the integral becomes,

$\int \frac{dx}{(1+x^4)^{1/4}} =\int \frac{1}{(1+\tan^2 \theta)^{1/4}} \frac{\sec^2 \theta d\theta}{2\sqrt{\tan \theta}} \\ \int \frac{dx}{(1+x^4)^{1/4}} =\int \frac{1}{\sqrt{\sec \theta}} \frac{\sec^2 \theta d\theta}{2\sqrt{\tan \theta}} \\ \int \frac{dx}{(1+x^4)^{1/4}} =\int \frac{\sec \theta d\theta}{2\sqrt{\sin \theta}} \\ \int \frac{dx}{(1+x^4)^{1/4}} =\int \frac{ \cos \theta d\theta}{2\cos^2 \theta \sqrt{\sin \theta}}$

$\int \frac{dx}{(1+x^4)^{1/4}} =\int \frac{ \cos \theta d\theta}{2(1-\sin^2 \theta) \sqrt{\sin \theta}}$

Make the substitution, $\sin \theta =t^2\\ \sin \theta d \theta=2tdt$

$\int \frac{dx}{(1+x^4)^{1/4}} =\int \frac{ 2tdt}{2(1-t^4)t}\\ \int \frac{dx}{(1+x^4)^{1/4}} =\int \frac{ dt}{2(1-t^4)}\\ \int \frac{dx}{(1+x^4)^{1/4}} =\frac{1}{2}\int (\frac{1}{1+t^2}+\frac{1}{1-t^2}) dt\\ \int \frac{dx}{(1+x^4)^{1/4}} =\frac{1}{2}\int \frac{1}{1+t^2} dt+\int \frac{1}{2}\frac{1}{1-t^2} dt\\ \int \frac{dx}{(1+x^4)^{1/4}} =\frac{1}{2}\tan ^{-1} (t)+\frac{1}{4}\ln |\frac{1+t}{1-t}| +C$

Note that $t= \sqrt{\sin \theta}=\frac{x}{(1+x^4)^{1/4}}$

Back substituting,

$\int \frac{dx}{(1+x^4)^{1/4}} =\frac{1}{2}\tan ^{-1} (\frac{x}{(1+x^4)^{1/4}} )+\frac{1}{4}\ln |\frac{1+\frac{x}{(1+x^4)^{1/4}} }{1-\frac{x}{(1+x^4)^{1/4}} }| +C \\ \int \frac{dx}{(1+x^4)^{1/4}} =\frac{1}{2}\tan ^{-1} (\frac{x}{(1+x^4)^{1/4}} )+\frac{1}{4}\ln |\frac{(1+x^4)^{1/4}+x }{(1+x^4)^{1/4}-x}| +C$