Mathematics

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

$0$

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

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