The value of $\int \frac{1}{x^2\left(x^4+1\right)^{3 / 4}} d x$, is

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

We have,

$I =\int \frac{1}{x^2\left(x^4+1\right)^{3 / 4}} d x$

$\Rightarrow I =\int \frac{1}{x^5\left(1+\frac{1}{x^4}\right)^{3 / 4}} d x$

$\Rightarrow I =-\frac{1}{4} \int\left(1+\frac{1}{x^4}\right)^{-3 / 4}\left(\frac{-4}{x^5}\right) d x$

$\Rightarrow I=-\frac{1}{4} \int\left(1+\frac{1}{x^4}\right)^{-3 / 4} d\left(1+\frac{1}{x^4}\right)$

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