Answer:
∫ x²Lnx dx = 1/3 [ x³Lnx - (1/3)x³] + C
Step-by-step explanation:
∫ x²Lnx dx
Integration by parts:
if we have u*v then D(u*v) = v*du + u*dv (1)
We make changes of variables :
Lnx dx = du then u = xLnx - x
v = x² then dv = 2xdx
And
∫ x²Lnx dx becomes ∫vdu
According to expression (1)
∫vdu = u*v - ∫udv
Now by substitution
∫vdu = x² ( xLnx - x ) - ∫( xLnx - x) 2xdx
∫ x²Lnx dx = x² ( xLnx - x ) - ∫ 2x²Lnxdx + ∫2x²dx
∫ x²Lnx dx = x² ( xLnx - x ) - 2 ∫x²Lnxdx + 2 (x³/3) + C
∫ x²Lnx dx + 2 ∫x²Lnxdx = x² ( xLnx - x ) + 2 (x³/3) + C
3 ∫ x²Lnx dx = x³Lnx -x³ + 2/3)x³ +C
∫ x²Lnx dx = 1/3 [ x³Lnx - (1/3)x³] + C
