How do you integrate #int x^2lnx# by parts from #[1,2]#? Calculus Techniques of Integration Integration by Parts 1 Answer Andrea S. Dec 28, 2016 #int_1^2x^2lnxdx=8/3ln2-7/9# Explanation: #int_1^2x^2lnxdx=int_1^2lnx d(x^3/3) = [x^3/3lnx]_1^2 - int_1^2x^3/3d(lnx) = 8/3ln2-int_1^2x^3/3(dx)/x=8/3ln2-int_1^2x^2/3dx=8/3ln2-[x^3/9]_1^2=8/3ln2-7/9# 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 1432 views around the world You can reuse this answer Creative Commons License