How do you integrate #int ln 2x^2 dx # using integration by parts? Calculus Techniques of Integration Integration by Parts 1 Answer Eddie Jul 12, 2016 reading this as : #int \ ln(2x^2) \ dx# then # = x ( ln2x^2 - 2) + C# Explanation: #int \ ln(2x^2) \ dx# using IBP #= int \d/dx(x) * ln2x^2 \ dx# # = x * ln2x^2 - int \ x * d/dx ( ln2x^2) \ dx# # = x ln2x^2 - int \ x * 1/(2x^2) 4x \ dx# # = x ln2x^2 - 2 int \ dx# # = x ln2x^2 - 2x + C# # = x ( ln2x^2 - 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 5427 views around the world You can reuse this answer Creative Commons License