## Calculus 8th Edition

$\frac{1}{18}$
$\int_0^1\frac{x}{(2x+1)^3}dx$ Let $u=2x+1$. Then $du=2\ dx$, so $dx=\frac{1}{2}\ du$. Also, since $2x=u-1$, $x=\frac{u-1}{2}$. $\int_{2*0+1}^{2*1+1}\frac{\frac{u-1}{2}}{u^3}*\frac{1}{2}du$ $=\frac{1}{4}\int_1^3\frac{u-1}{u^3}du$ $=\frac{1}{4}\int_1^3(\frac{u}{u^3}-\frac{1}{u^3})du$ $=\frac{1}{4}\int_1^3(u^{-2}-u^{-3})du$ $=\frac{1}{4}(\frac{u^{-1}}{-1}-\frac{u^{-2}}{-2})|_1^3$ $=\frac{1}{4}(-\frac{1}{u}-\frac{1}{-2u^2})|_1^3$ $=\frac{1}{4}((-\frac{1}{3}-\frac{1}{-2*3^2})-(-\frac{1}{1}-\frac{1}{-2*1^2}))$ $=\frac{1}{4}((-\frac{1}{3}-\frac{1}{-18})-(-1-\frac{1}{-2}))$ $=\frac{1}{4}(-\frac{5}{18}-(-\frac{1}{2}))$ $=\frac{1}{4}*\frac{2}{9}$ $=\boxed{\frac{1}{18}}$