Answer
$$
\int \frac{\cos ^{-1}\left(x^{-2}\right)}{x^{3}} d x =-\frac{1}{2} x^{-2} \cos ^{-1}\left(x^{-2}\right)+\frac{1}{2} \sqrt{1-x^{-4}}+C
$$
Work Step by Step
$$
\int \frac{\cos ^{-1}\left(x^{-2}\right)}{x^{3}} d x
$$
If we make the substitution
$$
u=x^{-2}, \quad d u=-2 x^{-3} d x
$$
and we look at the Table of Integrals , we see that the closest entry is number $88:$
\begin{aligned}
\int \frac{\cos ^{-1}\left(x^{-2}\right)}{x^{3}} d x &=-\frac{1}{2} \int \cos ^{-1} u d u \quad\left[\begin{array}{c}
u=x^{-2}, \\
d u=-2 x^{-3} d x
\end{array}\right] \\
& \stackrel{88}{=}-\frac{1}{2}\left(u \cos ^{-1} u-\sqrt{1-u^{2}}\right)+C \\
&=-\frac{1}{2} x^{-2} \cos ^{-1}\left(x^{-2}\right)+\frac{1}{2} \sqrt{1-x^{-4}}+C
\end{aligned}
