## Calculus, 10th Edition (Anton)

$-\frac{1}{40(5x^4+2)^2}+C$
$\int\frac{x^3}{(5x^4+2)^3}dx$ Let $u=5x^4+2$. Then $du=20x^3dx$, so $x^3dx=\frac{du}{20}$. $=\int\frac{1}{u^3}*\frac{du}{20}$ $=\frac{1}{20}\int u^{-3}du$ $=\frac{1}{20}(\frac{u^{-2}}{-2}+C)$ $=-\frac{1}{40u^2}+C$ $=-\frac{1}{40(5x^4+2)^2}+C$