\[\tan^{- 1}e^{\tan^{- 1}x} + c] | MATH - INTEGRATION BY SUBSTITUTION | SolveeIt

$\tan^{- 1}e^{\tan^{- 1}x} + c$

$\int_{}^{}\frac{1}{x(\log x)^{2}}\mspace{6mu} dx =$

$\frac{1}{\log x} + c$

$- \frac{1}{\log x} + c$

${loglog}x + c$

Answer

$- \frac{1}{\log x} + c$

Explanation

$\sin x + c$

$\cos x + c$

Put $x + c$ , then it reduces to $x^{2} + c$ .

Question: \[\tan^{- 1}e^{\tan^{- 1}x} + c\]...