Answer
$ -\displaystyle \frac{1}{8}\cdot(4+1/x^{2})^{4}$
Work Step by Step
$I=\displaystyle \int\frac{(4+1/x^{2})^{3}}{x^{3}}dx$
We need a substitution such that $du$ has $x^{3}$ in the denominator.
Substitute: $\left[\begin{array}{llll}
u & =4+\dfrac{1}{x^{2}}=4+x^{-2} & & \\
& & & \\
du & =-2x^{-3}dx =-\dfrac{2dx}{x^{3}} & \Rightarrow & \dfrac{dx}{x^{3}}=-\dfrac{du}{2}
\end{array}\right]$
$I=\displaystyle \int\frac{(4+1/x^{2})^{3}}{x^{3}}dx=\int[(4+1/x^{2})^{3}](\frac{dx}{x^{3}})$
apply the substitution
$I=\displaystyle \int u^{3}(-\frac{du}{2})=-\frac{1}{2}\int u^{3}du$
power rule
$=-\displaystyle \frac{1}{2}\cdot\frac{u^{4}}{4}+C$
bring back the variable $x$
=$ -\displaystyle \frac{1}{8}\cdot(4+1/x^{2})^{4}$
