Evaluate $\int\frac{dx}{x(2-\ln x^2)}$

Question

How to evaluate $\displaystyle\int\frac{dx}{x(2-\ln x^2)}$?

I tried to use an operator, but before that I distributed so $\displaystyle\int\frac{dx}{(2x-x\ln x^2)}$ my operator was $-x\ln x^2$ am I supposed to change its sign? However, after doing this I got $\displaystyle\int\frac{-x\ln x^2}{(-2x^2\ln x^2+1)}$ then my $u$ became $(-2x^2\ln x^2+1)$ so I have a problem to integrate it here. I also did the product rule for differentiating. Am I wrong for using an operator? I'm doing advance lessons, I hope someone can help.

Answer 1

Hint :

Use substitution $u=\ln x$, you will get $$\int\frac{dx}{x(2-\ln^2x)}=\int\frac{du}{2-u^2}=\frac{1}{2\sqrt{2}}\int\frac{du}{\sqrt{2}-u}+\frac{1}{2\sqrt{2}}\int\frac{du}{\sqrt{2}+u}$$

Edit :

Assuming the problem is $$\int\frac{dx}{x(2-\ln x^2)}$$ then making substitution $u=\ln x$ we have $$\int\frac{dx}{x(2-\ln x^2)}=\int\frac{dx}{x(2-2\ln x)}=\frac{1}{2}\int\frac{du}{1-u}$$ The latter expression is trivial.

Answer 2

u=2-log x^2

du=-2dx/x

answer will be -(log|2-log x^2|)/2