\( \int\displaystyle\frac{\sqrt{9+(\log x)^2}}{x}dx \) is equal to: (where \( C \) is an arbitrary constant and consider \( \log_e x=\log x \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{2}\log x\sqrt{9+(\log x)^2}+\displaystyle\frac{9}{2}\log\left|\log x+\sqrt{9+(\log x)^2}\right|+C \)
The \( \frac{1}{x} \) in the integrand is the derivative of log x, so let
\( \displaystyle t=\log x \)
\( \displaystyle dt=\frac{1}{x}\,dx \)
\( \displaystyle \int\frac{\sqrt{9+(\log x)^2}}{x}\,dx=\int\sqrt{3^2+t^2}\,dt \)
By the standard result
\( \displaystyle \int\sqrt{a^2+t^2}\,dt=\frac{t}{2}\sqrt{a^2+t^2}+\frac{a^2}{2}\log\left|t+\sqrt{a^2+t^2}\right|+C \)
with a = 3:
\( \displaystyle \int\sqrt{9+t^2}\,dt=\frac{t}{2}\sqrt{9+t^2}+\frac{9}{2}\log\left|t+\sqrt{9+t^2}\right|+C \)
Putting back \( t=\log x \):
\( \displaystyle =\frac{1}{2}\log x\sqrt{9+(\log x)^2}+\frac{9}{2}\log\left|\log x+\sqrt{9+(\log x)^2}\right|+C \)