The equal to integral ∫01 log⁡(1−x+1+x)dx is

I=∫01 log⁡(1−x+1+x)dx=xlog⁡(1−x+1+x)]01−∫01 x11−x+1+x−121−x+121+xdx=log⁡(2)+12∫01 x1−x+1+x1+x−1−x1−x2dx=12log⁡2+12∫01 x(1+x)−(1−x)1+x+1−x−21−x21−x2dx=12log⁡2+12∫01 11−x2−1dx=12log⁡2+π2−1.

The equal to integral ∫01 log⁡(1−x+1+x)dx is

I=∫01 log⁡(1−x+1+x)dx=xlog⁡(1−x+1+x)]01−∫01 x11−x+1+x−121−x+121+xdx=log⁡(2)+12∫01 x1−x+1+x1+x−1−x1−x2dx=12log⁡2+12∫01 x(1+x)−(1−x)1+x+1−x−21−x21−x2dx=12log⁡2+12∫01 11−x2−1dx=12log⁡2+π2−1.

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The equal to integral $\int_{0}^{1} \log (\sqrt{1 - x} + \sqrt{1 + x}) d x$ is

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$\frac{1}{2} (\log 2 - \frac{1}{2} + \frac{π}{4})$

$\frac{1}{2} (\log 2 - 1 + \frac{π}{2})$

$\frac{1}{3} (\log 4 - 1 + \frac{π}{8})$

$\frac{1}{4} (\log 3 - 1 + \frac{π}{2})$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int_{0}^{1} \log (\sqrt{1 - x} + \sqrt{1 + x}) d x \\ = x \log (\sqrt{1 - x} + \sqrt{1 + x})]_{0}^{1} \\ - \int_{0}^{1} x (\frac{1}{\sqrt{1 - x} + \sqrt{1 + x}}) (\frac{- 1}{2 \sqrt{1 - x}} + \frac{1}{2 \sqrt{1 + x}}) d x \\ = \log (\sqrt{2}) \\ + \frac{1}{2} \int_{0}^{1} \frac{x}{\sqrt{1 - x} + \sqrt{1 + x}} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 - x^{2}}}) d x \\ = \frac{1}{2} \log 2 \end{array}$

$\begin{array}{l} + \frac{1}{2} \int_{0}^{1} \frac{x}{(1 + x) - (1 - x)} \frac{1 + x + 1 - x - 2 \sqrt{1 - x^{2}}}{\sqrt{1 - x^{2}}} d x \\ = \frac{1}{2} \log 2 + \frac{1}{2} \int_{0}^{1} (\frac{1}{\sqrt{1 - x^{2}}} - 1) d x \\ = \frac{1}{2} (\log 2 + \frac{π}{2} - 1) . \end{array}$