Question

Solve :
$\int \frac{1}{x^2(x^4+1{)}^{3/4}}dx$

Answer 1

$\int \frac{dx}{x^2{(x^4+1)}^{\frac{3}{4}}}$

$=\int \frac{dx}{x^2.x^3{(1+\frac{1}{x^4})}^{\frac{3}{4}}}$...............taking $x^4$ common from under bracket term

Put $t=1+\frac{1}{x^4}\Rightarrow dt=-\frac{4}{x^5}dx$

so,integral is

$I=-\frac{1}{4}\int \frac{dt}{t^{\frac{3}{4}}}$

$=-\frac{1}{4}\left[\frac{t^{\frac{-3}{4}}+1}{\frac{-3}{4}+1}\right]+c$

$=-\frac{1}{4}\frac{t^{\frac{1}{4}}}{\frac{1}{4}}+c$

$=-t^{\frac{1}{4}}+c$

$=-{(1+\frac{1}{x^4})}^{\frac{1}{4}}+c$

$=-{\left(\frac{x^4}{1+x^4}\right)}^{\frac{1}{4}}+c$

$=-\frac{x}{\sqrt[4]{4x^4+1}}+c$