Answer
$-\displaystyle \frac{1}{50(5x^{2}-3)^{5}}+C$
Work Step by Step
$\displaystyle \int\frac{x}{(5x^{2}-3)^{6}}dx=\int x(5\mathrm{x}^{2}-3)^{-6}dx=$
Shortcut to apply:
$\displaystyle \quad\int g\cdot u^{n}dx=\frac{g}{u'}\cdot\frac{u^{n+1}}{n+1}+C \quad ($if $n\neq-1)$
$\left[\begin{array}{lll}
g(x)=x, & u(x)=5x^{2}-3, & n=-6\\
& u'(x)=10x &
\end{array}\right]$
$ =\displaystyle \frac{x}{10x}\cdot\frac{(5x^{2}-3)^{-5}}{-5}+C$
$=\displaystyle \frac{(5x^{2}-3)^{-5}}{-50}+C$
= $-\displaystyle \frac{1}{50(5x^{2}-3)^{5}}+C$
