How do you integrate #intdx/(x+xlnx^2)#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Jim H Apr 27, 2015 We know #d/dx(lnx) = 1/x#, so we should think about that when we see an integral involving #lnx#. #intdx/(x+xlnx^2) = int 1/(1+lnx^2) 1/x dx# #d/dx(1+lnx^2) = d/dx(lnx^2) = d/dx(2lnx) = 2/x# Integrate by substitution with #u = 1+2lnx# (Or #u = 1+lnx^2#, they are equal). Answer link Related questions How do you evaluate the integral #inte^(4x) dx#? How do you evaluate the integral #inte^(-x) dx#? How do you evaluate the integral #int3^(x) dx#? How do you evaluate the integral #int3e^(x)-5e^(2x) dx#? How do you evaluate the integral #int10^(-x) dx#? What is the integral of #e^(x^3)#? What is the integral of #e^(0.5x)#? What is the integral of #e^(2x)#? What is the integral of #e^(7x)#? What is the integral of #2e^(2x)#? See all questions in Integrals of Exponential Functions Impact of this question 1572 views around the world You can reuse this answer Creative Commons License