Question

The function $f:(0, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ be defined as, $f( x )=\arctan (\ln x )$, then the value of the integral $\int\limits_1^{ e } \frac{f( x )}{ x } dx$ 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)

$I =\int\limits_1^{ e } \frac{\tan ^{-1}(\ln x )}{ x } dx =\int\limits_0^1 \tan ^{-1}( t ) dt \quad($ where $\ln x = t )$ using I.B.P., we get
$\left.=\left.\tan ^{-1}( t )\right|_0 ^1-\int\limits_0^1 \frac{ t }{1+ t ^2} dt =\frac{\pi}{4}-\frac{1}{2} \ln \left(1+ t ^2\right)\right]_0^1=\frac{\pi}{4}-\frac{1}{2} \ln 2$