The value of $\int_{- 2}^{2}{\left\lbrack p\log{\left( \frac{1 + x}{1 - x} \right) + q\log{\left( \frac{1 - x}{1 + x} \right)^{- 2} + r}} \right\rbrack dx}$ depends on

Question

Dr. Abubacker · Accepted Answer

(c)
We have,
$I = \int_{- 2}^{2}{\left\lbrack p\log\left( \frac{1 + x}{1 - x} ight) + q\log\ \left( \frac{1 - x}{1 + x} ight)^{- 2} + r ightbrack dx}$
Since, $\log\ \left( \frac{1 + x}{1 - x} ight)$ is an odd function
$	herefore\int_{- 2}^{2}{\log{\left( \frac{1 + x}{1 - x} ight)dx = 0}}$
$\Rightarrow I = \int_{- 2}^{2}{rdx = r(2 + 2) = 4r}$

Q. The value of $\int_{- 2}^{2}{\left\lbrack p\log{\left( \frac{1 + x}{1 - x} \right) + q\log{\left( \frac{1 - x}{1 + x} \right)^{- 2} + r}} \right\rbrack dx}$ depends on

Explanation

Reviewed by Dr. Abubacker

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