The solution is shown here (sometimes Wolfram Alpha has trouble computing integrals with lnx properly, so I gave the backwards check).
I would start with a u-substitution and separate the integral.
Let:
u = x+1
du = dx
x = u - 1
=> int (u-1)lnudu
= int u lnudu - int lnudu
With these two integrals in mind, we can do Integration by Parts (assuming you already know the integral of lnx). Ignoring the int lnu, let:
s = lnu
ds = 1/udu
dt = udu
t = u^2/2
Thus:
st - int tds
= [(u^2lnu)/2 - int u^2/2*1/udu] - int lnudu
= (u^2lnu)/2 - 1/2int udu - int lnudu
= (u^2lnu)/2 - u^2/4 - (u lnu - u)
= (u^2lnu)/2 - u^2/4 - u lnu + u
= (u^2lnu)/2 - u lnu - u^2/4 + u
= ((x+1)^2ln(x+1))/2 - (x+1) ln(x+1) - (x+1)^2/4 + x + 1 + C
And 1 gets embedded into C:
= color(blue)(((x+1)^2ln(x+1))/2 - (x+1) ln(x+1) - (x+1)^2/4 + x + C)
