How do you integrate #(lnx)^2/x#? Calculus Techniques of Integration Integration by Parts 1 Answer Konstantinos Michailidis Jun 19, 2016 Notice that #(lnx)'=1/x# hence #int (lnx)^2/x dx=int (lnx)^2*(lnx)' dx=1/3*(lnx)^3+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 1460 views around the world You can reuse this answer Creative Commons License