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 1657 views around the world You can reuse this answer Creative Commons License