Question

Integral of $\frac{1}{1+(logx{)}^2}$ with respect to log x

• A. $\frac{ta{n}^{-1}\left(logx\right)}{x}+C$
• B. $ta{n}^{-1}\left(logx\right)+C$
• C. $\frac{ta{n}^{-1}x}{x}+C$
• D. None of these

Answer 1

The correct option is B $ta{n}^{-1}\left(logx\right)+C$

$\int \frac{1}{1+(logx{)}^2}d\left(logx\right)=\int \frac{dt}{1+t^2}[Putlogx=t\Rightarrow d(logx)=dt]=ta{n}^{-1}t+C=ta{n}^{-1}\left(logx\right)+C$