How do you integrate # ln(sqrt(x)) #? Calculus Techniques of Integration Integration by Parts 1 Answer Eddie Jul 12, 2016 #= 1/2 x( ln x - 1) + C# Explanation: #int \ ln(sqrt(x)) \ dx# using IBP: #int uv' = uv - int u'v# #= int \ ln(sqrt(x)) * (x)' \ dx# #= ln(sqrt(x)) * x - int (ln(sqrt(x)))' * x\ dx# #= 1/2 x ln x - int \ 1/(sqrt(x)) (1/2 * 1/sqrtx) * x\ dx# #= 1/2 x ln x - 1/2 int \ dx # #= 1/2 x ln x - 1/2 x + C# #= 1/2 x( ln x - 1) + 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 6083 views around the world You can reuse this answer Creative Commons License