∫01 ln⁡(1+x)1+x2dx=

Let I=∫01 ln⁡(1+x)1+x2dx Let x=tan⁡θ⇒dx=sec2⁡θdθI=∫0π/4 ln⁡(1+tan⁡θ)sec2⁡θsec2⁡θdθ=∫0π/4 ln⁡(1+tan⁡θ)dθ=∫0π/4 ln⁡1+tan⁡π4dθ=∫0π/4 ln⁡1+1−tan⁡θ1+tan⁡θdθ=∫0π/4 ln⁡21+tan⁡θdθ=∫0π/4 ln⁡2dθ−=∫0π4 ln⁡(1+tanθ )dθ⇒2I=π4ln⁡2⇒I=π8ln⁡2x=tan⁡θ;I⇒∫0π/4 ln⁡(1+tan⁡θ)sec2⁡θsec2⁡θdθ=π8ln⁡2

∫01 ln⁡(1+x)1+x2dx=

Let I=∫01 ln⁡(1+x)1+x2dx Let x=tan⁡θ⇒dx=sec2⁡θdθI=∫0π/4 ln⁡(1+tan⁡θ)sec2⁡θsec2⁡θdθ=∫0π/4 ln⁡(1+tan⁡θ)dθ=∫0π/4 ln⁡1+tan⁡π4dθ=∫0π/4 ln⁡1+1−tan⁡θ1+tan⁡θdθ=∫0π/4 ln⁡21+tan⁡θdθ=∫0π/4 ln⁡2dθ−=∫0π4 ln⁡(1+tanθ )dθ⇒2I=π4ln⁡2⇒I=π8ln⁡2x=tan⁡θ;I⇒∫0π/4 ln⁡(1+tan⁡θ)sec2⁡θsec2⁡θdθ=π8ln⁡2

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$\int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} dx =$

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$\frac{π}{4}$ In 2

$\frac{π}{2}$ In 2

$\frac{π}{8}$ In 2

$π$ In 2

answer is C.

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Detailed Solution

$L e t I = \int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} dx$
$Let x = \tan θ \Rightarrow d x = \sec^{2} θ d θ$
$\begin{array}{l} I = \int_{0}^{π / 4} \frac{\ln (1 + \tan θ)}{\sec^{2} θ} \sec^{2} θ d θ \\ = \int_{0}^{π / 4} \ln (1 + \tan θ) d θ \\ = \int_{0}^{π / 4} \ln (1 + \tan (\frac{π}{4})) d θ \\ = \int_{0}^{π / 4} \ln (1 + \frac{1 - \tan θ}{1 + \tan θ}) d θ \\ = \int_{0}^{π / 4} \ln (\frac{2}{1 + \tan θ}) d θ \\ = \int_{0}^{π / 4} \ln 2 d θ - = \int_{0}^{π 4} \ln (1 + \tan θ) d θ \\ \Rightarrow 2 I = \frac{π}{4} \ln 2 \\ \Rightarrow I = \frac{π}{8} \ln 2 \end{array}$
$x = \tan θ; I \Rightarrow \int_{0}^{π / 4} \frac{\ln (1 + \tan θ)}{\sec^{2} θ} \sec^{2} θdθ = \frac{π}{8} \ln 2$