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

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).

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