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The value of int0^1 (8log(1+x))/(1+x^2) ...

The value of `int_0^1 (8log(1+x))/(1+x^2) dx` is

A

`log 2`

B

`pi log 2`

C

`(pi)/8 log 2`

D

`(pi)/2 log 2`

Text Solution

Verified by Experts

The correct Answer is:
B
We have
`I=int_(0)^(1)(8log(1+x))/(1+x^(2))dx`
Put `x=tan theta`
`impliesI=int_(0)^((pi)/4) 8 . (log (1+tan theta))/(sec^(2) theta) sec^(2) theta d theta`
`=8 int_(0)^(pi//4) log (1+tan theta) d theta`
`=8 int_(0)^(pi//4) log (1+tan (pi//4-theta) d theta`
`=8int_(0)^(pi//4) log(1+(1-tan theta)/(1+tan theta)) d theta`
`=8int_(0)^(pi//4) log (2/(1+ tan theta)) d theta`
`=8(log2). (pi)/4-I`
`implies2I=2pi log 2`
`implies I=pi log 2`
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