Question

The function $f:(0,\infty )\to \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ be defined as, $f(x)=\arctan (\ln x)$, then the value of the integral $\underset{1}{\overset{e}{\int }}\frac{f(x)}{x}dx$ is equal to

• A. $\frac{\pi }{4}-\ln 2$
• B. $\frac{\pi }{4}-\frac{1}{2}\ln 2$
• C. $\frac{\pi }{4}+\ln 2$
• D. $\pi -\frac{1}{2}\ln 2$

Answer 1

Answer: (B)

$I=\underset{1}{\overset{e}{\int }}\frac{{\tan }^{-1}(\ln x)}{x}dx=\underset{0}{\overset{1}{\int }}{\tan }^{-1}(t)dt($ where $\ln x=t)$ using I.B.P., we get
${={{\tan }^{-1}(t)|}_0^1-\underset{0}{\overset{1}{\int }}\frac{t}{1+t^2}dt=\frac{\pi }{4}-\frac{1}{2}\ln \left(1+t^2\right)]}_0^1=\frac{\pi }{4}-\frac{1}{2}\ln 2$