I know I have to make a u substitution and then do integration by parts. $\int x\ln(1+x)dx$
$ u = 1 + x$
$du = dx$
$\int (u-1)(\ln u)du$
$\int u \ln u du - \int \ln u du$
I will solve the $\ln u$ problem first since it will be easier
$ \int \ln u du$
$u = \ln u$
$du = 1/u$
$dz = du$
$z = u$
$-(u\ln u - u)$
Now I will do the other part.
$\int u \ln u du$
$u = \ln u$ $du = 1/u$
$dz = udu$ $z = u^2 / 2$
$\frac {u^2 \ln u}{2} - \int u/2$
$\frac {u^2 \ln u}{2} - \frac{1}{2} \int u$
$\frac {u^2 \ln u}{2} - \frac{u^2}{2} $
Now add the other part.
$\frac {u^2 \ln u}{2} - \frac{u^2}{2} -u\ln u + u $
Now put u back in terms of x.
$\frac {(1+x)^2 \ln (1+x)}{2} - \frac{(1+x)^2}{2} -(1+x)\ln (1+x) + (1+x) $
This is wrong and I am not sure why.