How do you integrate #xln(1+x) dx#?

Question
Answer 1

The solution is shown here (sometimes Wolfram Alpha has trouble computing integrals with #lnx# properly, so I gave the backwards check).

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#

#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)#

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