How do you integrate #int sqrtx ln 2xdx#? Calculus Techniques of Integration Integration by Parts 1 Answer Tom Apr 12, 2015 By part : #du = sqrt(x) # #u = 2/3sqrt(x^3)# #v = ln(2x)# #dv = 1/x# #=[2/3sqrt(x^3)*ln(2x)]-2/3intsqrt(x^3)/xdx# Note : #sqrt(x^3)/x = x^(3/2-1)=x^(1/2) = sqrt(x)# Finally we have : #=[2/3sqrt(x^3)*ln(2x)]-4/9[sqrt(x^3)]+C# #=2/9sqrt(x^3)(3ln(2x)-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 1525 views around the world You can reuse this answer Creative Commons License