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.