∫∞0 log(1+x2)1+x2dx=

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Question

\int_{0}^{\infty} \frac{l o g (1 + x^{2})}{1 + x^{2}} d x =

Options:

A . π log 12

B . π log 2

C . 2π log 12

D . 2π log 2

Answer: Option B
:
B

L e t I = \int_{0}^{\infty} \frac{l o g (1 + x^{2})}{1 + x^{2}} d x

P u t x = t a n θ \Rightarrow d x = s e c^{2} θ d θ,

∴ I = \int_{0}^{\frac{n}{2}} l o g (s e c θ)^{2} d θ = 2 \int_{0}^{\frac{n}{2}} l o g s e c θ d θ

= - 2 \int_{0}^{\frac{n}{2}} l o g c o s θ d θ

= - 2. \frac{π}{2} l o g \frac{1}{2} = - π l o g \frac{1}{2} = π l o g 2

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