Question

Encontre a primitiva da integral x^2 Lnx dx,sendo u=u=lnx e dv=x^2 dx

Answer 1

Olá. 

U= lnx derivando U -  du/dx = 1/x   du = dx/x

dv= x^2dx  integrando dv = x^n +1/n+1 portanto dv= x^2dx 

V= x^3/3 

∫u.v'dx = u.v - ∫vdu
∫lnx. x^2dx = lnx.x³/3 - ∫ x³/3dx/x 
∫lnx. x^2dx  =  x³lnx/3 -  1/3 . ∫ x²dx
∫lnx. x^2dx =  x³lnx/3 - 1/3 . x³/3
∫lnx. x^2dx = x³lnx/3 -  x³/9
∫lnx. x^2dx = (3x³lnx - X³)/9
∫lnx. x^2dx = X³(3lnx - 1)/9

espero ter ajudado.