Integration by $u$ substitution

Let $u=x+1$. Then $du=dx$ and we have
\[\int_0^1 \frac{x^2}{(x+1)^3}dx=\int_1^{2}\frac{(u-1)^2}{u^3}du=\int_1^{2}\frac{u^2-2u+1}{u^3}du\]
\[=\int_1^{2}\frac{1}{u}-2u^{-2}+u^{-3}du=(\ln (u)+2u^{-1}+\frac{u^{-2}}{-2})\big|_1^{2}\]
\[=(\ln (u)+\frac{2}{u}-\frac{1}{2u^2})\big|_1^{2}\]
\[=(\ln 2+1-\frac{1}{2\times 4})-(\ln 1+2-\frac{1}{2})\]
\[=\ln 2-1-\frac{1}{8}+\frac{1}{2}=\ln2+\frac{-8-1+4}{8}\]
\[=\ln 2-\frac{5}{8}.\]

Note that $\ln1 =0$.