How do you integrate #ln(x^(1/3))#? Calculus Techniques of Integration Integration by Parts 1 Answer sjc Mar 19, 2018 #1/3xlnx-1/3x+c# Explanation: #I=intln(x^(1/3))dx# using the laws of logs #I=int 1/3lnxdx# # we will integrate by parts #I=1/3intlnxdx# #I=intu(dv)/(dx)dx=uv-intv(du)/(dx)dx# #u=lnx=>(du)/(dx)=1/x# #(dv)/(dx)=1=>v=x# #:.I=1/3[xlnx-intx xx 1/xdx]# #I=1/3[xlnx-intdx]# #=1/3[xlnx-x]+c# #1/3xlnx-1/3x+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 9085 views around the world You can reuse this answer Creative Commons License