How do you integrate #xlnx#? Calculus Techniques of Integration Integration by Parts 1 Answer Konstantinos Michailidis Oct 24, 2015 It is #int xlnxdx=int (x^2/2)'lnxdx=x^2/2*lnx-intx^2/2*lnx'dx= x^2/2*lnx-intx^2/2*1/xdx=x^2/2*lnx-intx/2dx=x^2/2*lnx-x^2/4+c# Finally #int xlnxdx=x^2/2*lnx-x^2/4+c# We used integration by parts #intf'(x)*g(x)dx=f(x)g(x)-intf(x)*g'(x)dx# 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 1663 views around the world You can reuse this answer Creative Commons License