How do you integrate #int sqrtx ln x dx # using integration by parts? Calculus Techniques of Integration Integration by Parts 1 Answer GiĆ³ Jun 22, 2016 I found: #2/3x^(3/2)[ln(x)-2/3]+c# Explanation: Try this: 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 4530 views around the world You can reuse this answer Creative Commons License