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The value of 018log1+x11+x2dx is

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a
log 2
b
πlog2
c
π8log⁡2
d
π2log⁡2

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detailed solution

Correct option is B

Let I=8∫01 log⁡(1+x)1+x2dx=8∫0π/4 log⁡(1+tan⁡θ)sec2⁡θsec2⁡θdθ(x=tan⁡θ)So I=8∫0π log⁡(1+tan⁡θ)dθ=8∫0π/4 log⁡1+tan⁡π4−θdθ∫0a f(x)dx=∫0a f(a−x)dx=8∫0π/4 log⁡1+1−tan⁡θ1+tan⁡θdθ=8∫0π/4 log⁡21+tan⁡θdθ=(8log⁡2)π4−I⇒ 2I=(8log⁡2)π4⇒I=πlog⁡2


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