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

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

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

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

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

$\Rightarrow I=-\frac{1}{4} \int \frac{d t}{t^{3 / 4}}=-\frac{1}{4} \int t^{-3 / 4} d t=-\frac{1}{4} \times 4 t^{1 / 4}+c$

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

Hence (2) is the correct answer.