∫ log(1 + 1/x) 1/(1 + x^2)dx, x ∈ [0, ∞]= A. π ln 2 B. –π ln 2 C. 0 D. -π/2 In 2

Question

\(\int\limits_{0}^{\infty} \)log\(\Big(\text x+\cfrac{1}{\text x }\Big)\cfrac{1}{1+\text x^2}d\text x=\)

A. π ln 2

B. –π ln 2

C. 0

D. -π/2 In 2

Answer 1

Correct option is A. π ln 2

Given,

 \(\int\limits_{0}^{\infty} \)log\(\Big(\text x+\cfrac{1}{\text x }\Big)\cfrac{1}{1+\text x^2}d\text x\)

Put x = tan θ

dx = sec² θ d θ

Limits also will be changed accordingly,

x = 0 → θ = 0

x = ∞ → θ = \(\cfrac{\pi}2\)

(Some standard notations which we need to remember)