The integral $\int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ equals ________.

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

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

$\int \frac{dx}{x^2\left(x^4+1\right)^{\frac{3}{4}}}=\int\frac{x^{-5}}{(1+x^{-4})^{\frac{3}{4}}}dx$

$y=1+x^{-4}⇒dy=-4x^{-5}dx$

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

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