How do you integrate # ln x / x^(1/2)#? Calculus Techniques of Integration Integration by Parts 1 Answer Eddie Jul 5, 2016 #= 2 sqrt x ln x - 4 sqrt x + C# Explanation: Use IBP #int u v' = uv - int u'v# here #u = ln x, u' = 1/x# #v' = x^{-1/2}, v = 2 x^{1/2}#, using the power rule so we have #2 sqrt x ln x - 2 int dx qquad 1/x sqrt x# #= 2 sqrt x ln x - 2 int dx qquad x^{-1/2}# #= 2 sqrt x ln x - 2*x^{1/2}/(1/2) + C# #= 2 sqrt x ln x - 4 sqrt x + 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 5482 views around the world You can reuse this answer Creative Commons License