Integral of $\frac{1}{1+(\log x)^2}$ w.r.t. log x is:
Answer & explanation
Correct answer: option 2
$\frac{1}{1+(\log x)^2}d(\log x)=\int\frac{dt}{1-t^2}$ [Putting log x = t ⇒ $d(\log x)=dt=\tan^{-1}t+C=\tan^{-1}(\log x)+C$
Integral of $\frac{1}{1+(\log x)^2}$ w.r.t. log x is:
Correct answer: option 2
$\frac{1}{1+(\log x)^2}d(\log x)=\int\frac{dt}{1-t^2}$ [Putting log x = t ⇒ $d(\log x)=dt=\tan^{-1}t+C=\tan^{-1}(\log x)+C$