Question

If f(x) =$\int_{1}^{x}\frac{\tan^{- 1}(t)}{t}$dt"xĪR⁺ then the value of f(e²) – f$\left( \frac{1}{e^{2}} \right)$ is

• A. 0
• B. $\frac{\pi}{2}$
• C. p
• D. 2p

Answer 1

Answer: (C)

p

Answer 2

f(x) Ž $\int_{1}^{x}\frac{\tan^{- 1}(t)}{t}$ dt

f$\left( \frac{1}{x} \right)$ Ž $\int_{1}^{1/x}\frac{\tan^{- 1}(t)}{t}$ dt

put t = 1/u

dt = –1/u² du

f(1/x) Ž =$\int_{1}^{x}\frac{\tan^{- 1}\left( \frac{1}{u} \right)}{1/u}$ $\left( - \frac{1}{u^{2}} \right)$du

f(1/x) ŗ – $\int_{1}^{x}\frac{\tan^{- 1}\left( \frac{1}{u} \right)}{u}$= – $\int_{1}^{x}\frac{\cot^{- 1}(u)}{u}$ du

= – $\int_{1}^{x}\frac{\cot^{- 1}(t)}{t}$dt

f(x) – f(1/x) = $\int_{1}^{x}\frac{\tan^{- 1}t + \cot^{- 1}t}{t}$ dt

= $\int_{1}^{x}{\frac{\pi}{2} \times \frac{1}{t}}$ dt

f(x) – f(1/x) = $\frac{\pi}{2}$ log(x)

f(e²) – f(1/e²) = $\frac{\pi}{2}$ log_ee² = p