What is the integral of #ln(x^2)#?

1 Answer

It's #2x ln(x) -2x + C#

By integral I'm assuming it's meant the indefinite integral.

Using the fundamental proprety of logarithms that

#log_c(a^b)=b log_c(a)#

We get

#int ln(x^2) dx = int 2 ln(x) dx = 2 int ln(x) dx#

Integrating by parts, chosing #dv=1# and #u=ln(x)# we get

#int ln (x) dx = x ln(x) - int x/x dx = x ln(x) - int 1 dx = x ln(x) -x + C/2#

(we take the constant to be #C/2# for convinience, and we can do this because the integration constant is taken arbitrarily)

Therefore,

#int ln(x^2) dx = 2x ln(x) -2x + C#