Answer
$\dfrac{\ln^2 (x+1)}{2}\ln [\ln (x+1)] -\dfrac{\ln^2 (x+1)}{4}+C$
Work Step by Step
Consider that $a=x+1$ and $da=dx$
$\displaystyle \int (x+1) \ln (x+1) \ dx= \int a \ln a \ da$
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
Here, $u=\ln a$ and $dv=a \ da \implies v=\dfrac{a^2}{2}$
or, $=\dfrac{a^2}{2}\ln a -\dfrac{a^2}{4}+C$
Now, we will use back substitution $a=\ln (x+1)$
Therefore, we have:
$\int a \ln a \ da=\dfrac{\ln^2 (x+1)}{2}\ln [\ln (x+1)] -\dfrac{\ln^2 (x+1)}{4}+C$
