what is the integration of (x-1)lnx dx?

u = lnx
du = 1/x dx

dv = x-1
v = x^2 - x

uv - integral of vdu
(x^2 - x)lnx - integral of (x^2 - x)/x dx
(x^2 - x)lnx - integral of (x - 1) dx
x^2lnx - xlnx - x^2 +x +C

∫ (x-1)lnx dx=
let v=x-1
=∫ v*ln(v+1) dv=
=½∫ ln(v+1) d(v²)=
=½v²ln(v+1) - ½∫ v² d(ln(v+1))=
=½v²ln(v+1) - ½∫ v²/(v+1) dv=
=½v²ln(v+1) - ½∫ (v²-1+1)/(v+1) dv=
=½v²ln(v+1) - ½∫ (v²-1)/(v+1)+1/(v+1) dv=
=½v²ln(v+1) - ½∫ v-1+1/(v+1) dv=
=½v²ln(v+1) - ¼v² + ½v -½ln|v+1| +C=
=½(x-1)²ln(x-1+1) - ¼(x-1)² + ½(x-1) -½ln|x-1+1| +C=
=½(x-1)²ln(x) - ¼(x-1)² + ½(x-1) -½ln|x| +C

http://www.wolframalpha.com/input/?i=%28%C2%BD%28x-1%29%C2%B2ln%28x%29+-+%C2%BC%28x-1%29%C2%B2+%2B+%C2%BD%28x-1%29+-%C2%BDln|x|%29`