Answer
$\frac{1}{3}\left( {{{\left( {\ln 4} \right)}^3} - {{\left( {\ln 2} \right)}^3}} \right)=\dfrac{7\ln^3{2}}{3}$
Work Step by Step
$$\eqalign{
& \int_1^3 {\frac{{{{\left( {\ln \left( {v + 1} \right)} \right)}^2}}}{{v + 1}}} dv \cr
& {\text{Use substitution:}}\cr
&{\text{Let }}u = \ln \left( {v + 1} \right),{\text{ so that }}du = \frac{1}{{v + 1}}dv \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}v = 3,{\text{ then }}u = \ln \left( {3 + 1} \right) = \ln 4 \cr
& \,\,\,\,\,\,{\text{If }}v = 1,{\text{ then }}u = \ln \left( {1 + 1} \right) = \ln 2 \cr
& {\text{Then}} \cr
& \int_1^3 {\frac{{{{\left( {\ln \left( {v + 1} \right)} \right)}^2}}}{{v + 1}}} dv = \int_{\ln 2}^{\ln 4} {\frac{{{{\left( u \right)}^2}}}{{v + 1}}} \left( {v + 1} \right)du \cr
& = \int_{\ln 2}^{\ln 4} {{u^2}} du \cr
& {\text{integrate}} \cr
& = \left( {\frac{{{u^3}}}{3}} \right)_{\ln 2}^{\ln 4} \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{1}{3}\left( {{{\left( {\ln 4} \right)}^3} - {{\left( {\ln 2} \right)}^3}} \right) \cr} $$
