The value of $\int \frac{x^7}{\left(1-x^2\right)^5} d x$, is

$\frac{1}{8} \frac{x^8}{\left(1-x^2\right)^4}+C$

We have,

$I=\int \frac{x^7}{\left(1-x^2\right)^5} d x$

$\Rightarrow I =\int \frac{\sin ^7 \theta}{\left(1-\sin ^2 \theta\right)^5} \cos \theta d \theta$, where $x=\sin \theta$

$\Rightarrow I =\int \tan ^7 \theta \sec ^2 \theta d \theta$

$\Rightarrow I =\int(\tan \theta)^7 d(\tan \theta)=\frac{1}{8} \tan ^8 \theta+C$

$\Rightarrow I=\frac{1}{8} \frac{x^8}{\left(1-x^2\right)^4}+C$