$\int \frac{\sqrt{1+x^2}}{x^4} dx$ is equal to :

$-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3 / 2}+c$

Let $I=\int \frac{\sqrt{1+x^2}}{x^4} d x=\int \frac{\sqrt{1+\frac{1}{x^2}}}{x^3} d x$

Let $1+\frac{1}{x^2}=t \Rightarrow -\frac{2}{x^3} dx=dt$

$\Rightarrow I=-\frac{1}{2} \int \sqrt{t} dt$

$=-\frac{1}{2} . \frac{2}{3} t^{3 / 2}+c$

$=-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3 / 2}+c$

Hence (3) is the correct answer.