Answer
$$\int^{3}_1\frac{(\ln(v+1))^2}{v+1}dv=\frac{(\ln4)^3-(\ln2)^3}{3}=\frac{7 (\ln{2})^3}{3}$$
Work Step by Step
$$A=\int^{3}_1\frac{(\ln(v+1))^2}{v+1}dv$$
We set $u=\ln(v+1)$, which means $$du=\frac{(v+1)'}{v+1}dv=\frac{1}{v+1}dv$$
For $v=1$, we have $u=\ln2$
For $v=3$, we have $u=\ln4$
Therefore, $$A=\int^{\ln4}_{\ln2}u^2du$$ $$A=\frac{1}{3}u^3\Big]^{\ln4}_{\ln2}$$ $$A=\frac{(\ln4)^3-(\ln2)^3}{3}=\frac{7 (\ln{2})^3}{3}$$
