How do you integrate intdx/(x+xlnx^2)?

1 Answer
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).