How do you find the integral of #6x ln x dx #? Calculus Techniques of Integration Integration by Parts 1 Answer GiĆ³ Jul 17, 2015 I found: #3x^2[ln(x)-1/2]+c# Explanation: I would try By Parts: #int6xln(x)dx=6x^2/2ln(x)-int6x^cancel(2)/2*1/cancel(x)dx=# #=3x^2ln(x)-int3xdx=# #=3x^2ln(x)-3x^2/2+c=# #=3x^2[ln(x)-1/2]+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 3802 views around the world You can reuse this answer Creative Commons License