Question

Integration of \[\frac{1}{1 + \left( \log_e x \right)^2}\] with respect to log_e x is

Options

• \[\frac{\tan^{- 1} \left( \log_e x \right)}{x} + C\]

• \[\tan^{- 1} \left( \log_e x \right) + C\]

• \[\frac{\tan^{- 1} x}{x} + C\]

• none of these

Answer 1

\[\tan^{- 1} \left( \log_e x \right) + C\]

\[\text{We have to integrate }\frac{1}{1 + \left( \log_e x \right)^2}\text{ with respect to }\log {}_e x \]
\[\text{Let }I = \int\frac{d \left( \log_e x \right)}{1 + \left( \log_e x \right)^2}\]
\[\text{Putting }\log_e x = t\]
\[d \left( \log_e x \right) = dt\]
\[ \therefore I = \int\frac{dt}{1 + t^2}\]
\[ = \tan^{- 1} \left( t \right) + C\]
\[ = \tan^{- 1} \left( \log_e x \right) + C ...............\left( \because t = \log_e x \right)\]