$\int \frac{1+\log x}{\sqrt{x^{2 x}-1}} d x$ equals
Answer & explanation
Correct answer: option 1
We have,
$I =\int \frac{1+\log x}{\sqrt{x^{2 x}-1}} d x$
$=\int \frac{1}{x^x \sqrt{\left(x^x\right)^2-1^2}} d\left(x^x\right)=\sec ^{-1}\left(x^x\right)+C$