CUET Preparation Today
$\int \frac{1+\log x}{\sqrt{x^{2 x}-1}} d x$ equals |
$\sec ^{-1}\left(x^x\right)+C$ $\log \left|x^x+\sqrt{x^{2 x}-1}\right|+C$ $\log \left|x^x-\sqrt{x^{2 x}-1}\right|+C$ none of these |
$\sec ^{-1}\left(x^x\right)+C$ |
We have, $I =\int \frac{1+\log x}{\sqrt{x^{2 x}-1}} d x$ $=\int \frac{1}{x^x \sqrt{\left(x^x\right)^2-1^2}} d\left(x^x\right)=\sec ^{-1}\left(x^x\right)+C$ |
