Integral of (x^2)lnx?

Question

Is there a way to integrate this function without using approximation? Thank you in advance for the answer.

Answer 1

let u=ln(x), dv=(x^2)dx
then du=dx/x and v=(1/3)x^3

by integration by parts

∫[(x^2)lnx]dx= ∫udv= uv-∫vdu= (1/3)x^3•ln(x)-1/3∫(x^2)dx = (1/3)x^3•ln(x) -(1/9)x^3

Answer 2

∫[(x^2)lnx]dx
= lnx∫x^2dx - ∫(d/dxlnx∫(x^2)dx)dx
= (x^3)*(1/3)*ln(x) -∫(1/x)*(x^3)*(1/3)dx
= (1/3)x^3*lnx-1/3∫(x^2)dx
= (1/3)*x^3*ln(x) -(1/9)*x^3 + c