The value of ∫log⁡(1−x)dx is

Integrating by parts taking log (1−x) as the first function, the given integral can be written as xlog⁡(1−x)+12∫x1−xdx=xlog⁡(1−x)+∫t21−tdt−xlog⁡(1−x)+∫−t−1+11−tdt=(x−1)log⁡(1−x)-x2−x+C.

The value of ∫log⁡(1−x)dx is

Integrating by parts taking log (1−x) as the first function, the given integral can be written as xlog⁡(1−x)+12∫x1−xdx=xlog⁡(1−x)+∫t21−tdt−xlog⁡(1−x)+∫−t−1+11−tdt=(x−1)log⁡(1−x)-x2−x+C.

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The value of $\int \log (1 - \sqrt{x}) d x$ is

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$x \log (1 - \sqrt{x}) - \frac{1}{2} x - \frac{3}{2} \sqrt{x} + C$

$(x - 1) \log (1 - \sqrt{x}) - \frac{1}{2} x - \sqrt{x} + C$

none of these

$x^{2} \log (1 - \sqrt{x}) - \frac{1}{2} \sqrt{x} + C$

answer is B.

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Detailed Solution

Integrating by parts taking log $(1 - \sqrt{x})$ as the first function, the given integral can be written as

$\begin{array}{l} x \log (1 - \sqrt{x}) + \frac{1}{2} \int \frac{\sqrt{x}}{1 - \sqrt{x}} d x \\ = x \log (1 - \sqrt{x}) + \int \frac{t^{2}}{1 - t} d t \\ - x \log (1 - \sqrt{x}) + \int (- t - 1 + \frac{1}{1 - t}) d t \\ = (x - 1) \log (1 - \sqrt{x}) - \frac{x}{2} - \sqrt{x} + C . \end{array}$