Integration by substitution Dx/ x2 (x4+1)power …

{tex}\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}{/tex}
{tex}\int \frac{1 d x}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}{/tex}
Taking x⁴ common from denominator
{tex}=\int \frac{1 d x}{x^{2}\left(x^{4}\right)^{\frac{3}{4}}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}{/tex}
{tex}=\int \frac{d x}{x^{2}\left(x^{3}\right)\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}{/tex}
{tex}=\int \frac{d x}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}{/tex}
{tex}\text { Let } t=1+\frac{1}{x^{4}}{/tex}
{tex}\frac{d t}{d x}=-\frac{4}{x^{5}}{/tex}
{tex}-\frac{d t}{4}=\frac{d x}{x^{5}}{/tex}
Substituting value of x and dx
{tex}=\frac{-1}{4} \int \frac{d t}{t^{\frac{3}{4}}}{/tex}
{tex}=\frac{-1}{4} \int t^{\frac{-3}{4}} d t{/tex}
{tex}=\frac{-1}{4} \int t^{\frac{-3}{4}} d t{/tex}
{tex}=\frac{-1}{4}\left[\frac{t^{\frac{-3}{4}}+1}{\frac{-3}{4}+1}\right]+C{/tex}
{tex}=-t^{\frac{1}{4}}+C {/tex}
{tex}=-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c{/tex}