Evaluating an integral using logarithms.

Question

Evaluate the integral:

$$ \int \frac{1}{x\log_3 x} dx $$

I tried to change it to this form :

$$ \int \frac{\ln 3}{x\ln x} dx $$

But i couldn't continue. How could i arrive to this form $ \dfrac{D(f(x))}{f(x)} $

Answer 1

$$ (\ln 3)\int \frac{1}{\ln x} \Big(\frac{dx}{x}\Big) = (\ln 3) \int \frac 1 u \, du. $$

Answer 2

$$I=\int \frac{1}{x\log_3x}dx$$$$=\int \frac{\ln3}{x \ln x}dx.$$Now let $\ln x=u \implies\frac{dx}{x}=du$.Therefore $I=\ln3 \int \frac{1}{u}du=\ln3(\ln u+c)=\ln3\ln(\ln x)+c'.$ Where $c'=c\ln3$