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 2402 views around the world You can reuse this answer Creative Commons License