How do you integrate #int (x+1)^2ln3x# by integration by parts method? Calculus Techniques of Integration Integration by Parts 1 Answer Alberto P. 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# Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 2253 views around the world You can reuse this answer Creative Commons License