What is the integral of #(x^2)(lnx)#? Calculus Techniques of Integration Integration by Parts 1 Answer Cem Sentin Apr 6, 2018 #int x^2*Lnx*dx=x^3/3*Lnx-x^3/9+C# Explanation: After setting #dv=x^2*dx# and #u=Lnx# for using integration by parts, #v=x^3/3# and #du=dx/x# Hence, #int udv=uv-int vdu# #int x^2*Lnx*dx=x^3/3*Lnx-int x^3/3*dx/x# =#x^3/3*Lnx-int x^2/3*dx# =#x^3/3*Lnx-x^3/9+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 65790 views around the world You can reuse this answer Creative Commons License