The value of _0^1 0.05em log ⁡ (1+x )1+x^2 dx is

Question

Accepted Answer

Correct answer: C. (c) : Let I= _0^1 0.05em log ⁡ (1+x )1+x^2 dx Put x= tan ⁡ ⇒ dx= sec^2 ⁡ θdθ When x=0 , =0 and when x=1 , = 4 ∴I= _0^ /4 log ⁡ (1+ tan ⁡ ) dθ (i) = _0^ /4 log ⁡(1+ tan ⁡( 4- )) dθ = _0^ /4 log ⁡(1+ 1- tan ⁡ 1+ tan ⁡ ) dθ ⇒ I= _0^ /4 ( log ⁡2- log ⁡ (1+ tan ⁡ ) ) dθ (ii) Adding (i) and (ii), we get 2I= _0^ /4 log ⁡2 dθ= log ⁡2 [ ]_0^ /4 ⇒ I= 12 log ⁡2 4= 8 log ⁡2

$The value of \int_{0}^{1} \frac{log ( 1 + x )}{1 + x ^{2}} dx is$

Solution

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