Mathematics

The value of $$\displaystyle\int_{-1}^{1}\log \left (\dfrac {x - 1}{x + 1}\right )dx$$ is


ANSWER

$$0$$


SOLUTION
Let $$I = \int_{-1}^{1} \log \left (\dfrac {x - 1}{x + 1}\right ) dx$$

Again, let $$f(x) = \log \left (\dfrac {x - 1}{x + 1}\right )$$

$$f(-x) = \log \left (\dfrac {-x - 1}{-x + 1}\right )$$

$$= \log \left (\dfrac {1 + x}{x - 1}\right )$$

$$= -\log \left (\dfrac {x - 1}{x + 1}\right )$$

$$= -f(x)$$

$$\Rightarrow f(x)$$ is an odd function.

$$\therefore I = 0$$
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Single Correct Medium Published on 17th 09, 2020
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