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)#
