$\int \frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}} dx$ is equal to:

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

The correct answer is Option (2) → $-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+C$

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

$=\int\frac{x^{-5}}{(1+x^{-4})^{\frac{3}{4}}}dx$

let $y=1+x^{-4}$

$dy=-4x^{-5}dx$

$⇒-\frac{1}{4}\int\frac{dy}{y^{\frac{3}{4}}}=-\frac{1}{4}\frac{y^{\frac{1}{4}}}{\frac{1}{4}}+C$

$=-y^{\frac{1}{4}}+C$

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