How do you integrate int (x+1)^2ln3x by integration by parts method?

1 Answer
Nov 3, 2016

(x+1)^3/3ln3x-1/3(x^3/3+3/2x^2+3x+lnabs(x))+c

Explanation:

f^'(x)=(x+1)^2 =>f(x)=(x+1)^3/3
g(x)=ln3x => g'(x)=1/x

int(x+1)^2ln3x\ dx=(x+1)^3/3ln3x-1/3int(x+1)^3\ 1/x\ dx=
=(x+1)^3/3ln3x-1/3int(x^3+3x^2+3x+1)/x\ dx=
=(x+1)^3/3ln3x-1/3intx^2+3x+3+1/x\ dx=
=(x+1)^3/3ln3x-1/3(x^3/3+3/2x^2+3x+lnabs(x))+c