Question

Integration of $\frac{1}{1+{\left({\log }_{e}x\right)}^2}$ with respect to log_e x is
(a) $\frac{{\tan }^{-1}\left({\log }_{e}x\right)}{x}+C$

(b) ${\tan }^{-1}\left({\log }_{e}x\right)+C$

(c) $\frac{{\tan }^{-1}x}{x}+C$

(d) none of these

Answer 1

(b) ${\tan }^{-1}\left({\log }_{e}x\right)+C$

$$\begin{gathered} \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{integrate}\frac{1}{1+{\left({\log }_{e}x\right)}^2}\mathrm{with}\mathrm{respect}\mathrm{to}\log {}_{e}x \\ \mathrm{Let}I=\int \frac{d\left({\log }_{e}x\right)}{1+{\left({\log }_{e}x\right)}^2} \\ \mathrm{Putting}{\log }_{e}x=t \\ d\left({\log }_{e}x\right)=dt \\ \therefore I=\int \frac{dt}{1+t^2} \\ ={\tan }^{-1}\left(t\right)+C \\ ={\tan }^{-1}\left({\log }_{e}x\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because \mathit{}t={\log }_{e}x\right) \end{gathered}$$