The value of ∫018log(1+x)1+x2dx is

I=∫018 log(1+x)1+x2dxPut x=tanθdx=Sec2θ dθ=∫0π/48 log(1+tanθ)1+tan2θSec2θ dθI=8∫log0π/41+tanπ4−θdθ=8∫0π/4log1+1−tanθ1+tanθdθ=8∫0π/4log(21+tanθ)dθI=8∫0π/4(log2)dθ−8∫0π/4log(1+tanθ)dθI=8 log2θ0π/4−I2I=8(log 2)π/4I=πlog2

The value of ∫018log(1+x)1+x2dx is

I=∫018 log(1+x)1+x2dxPut x=tanθdx=Sec2θ dθ=∫0π/48 log(1+tanθ)1+tan2θSec2θ dθI=8∫log0π/41+tanπ4−θdθ=8∫0π/4log1+1−tanθ1+tanθdθ=8∫0π/4log(21+tanθ)dθI=8∫0π/4(log2)dθ−8∫0π/4log(1+tanθ)dθI=8 log2θ0π/4−I2I=8(log 2)π/4I=πlog2

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The value of $\int_{0}^{1} \frac{8 \log (1 + x)}{1 + x^{2}} d x$ is

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$\frac{π}{8} \log 2$

$\frac{π}{2} \log 2$

log 2

$π \log 2$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int_{0}^{1} \frac{8 \log (1 + x)}{1 + x^{2}} d x \\ P u t x = \tan θ \\ d x = S e c^{2} θ d θ \\ = \int_{0}^{π / 4} \frac{8 \log (1 + \tan θ)}{1 + \tan^{2} θ} S e c^{2} θ d θ \\ I = 8 {\int \log}_{0}^{π / 4} (1 + \tan (\frac{π}{4} - θ)) d θ \\ = 8 \int_{0}^{π / 4} \log (1 + \frac{1 - \tan θ}{1 + \tan θ}) d θ \\ = 8 \int_{0}^{π / 4} \log (\frac{2}{1 + \tan θ}) d θ \\ I = 8 \int_{0}^{π / 4} (\log 2) d θ - 8 \int_{0}^{π / 4} \log (1 + \tan θ) d θ \\ I = 8 \log 2 {(θ)}_{0}^{π / 4} - I \\ 2 I = 8 (\log 2) π / 4 \\ I = π \log 2 \end{array}$