If $f(x) = \log\left( \frac{1 + x}{1 - x} \right)$, $- 1

Question

If $f(x) = \log\left( \frac{1 + x}{1 - x} \right)$, $- 1 < x < 1$, then $f\left( \frac{3x + x^{3}}{1 + 3x^{2}} \right) - f\left( \frac{2x}{1 + x^{2}} \right)$ is

Dr. Abubacker · Accepted Answer

(d)
$	herefore\ f\left( \frac{3x + x^{3}}{1 + 3x^{2}} ight) - f\left( \frac{2x}{1 + x^{2}} ight)$
$\ \ \ \ {= log}\left( \frac{1 + \left( \frac{3x + x^{3}\ }{1 + 3x^{2}} ight)}{1 - \left( \frac{3x + x^{3}}{1 + 3x^{2}} ight)} ight) - \log\left( \frac{1 + \frac{2x}{1 + x^{2}}}{1 - \frac{2x}{1 + x^{2}}} ight)$
$\ \ \ \  = \log{\left( \frac{1 + x}{1 - x} ight)^{3} - \log\left( \frac{1 + x}{1 - x} ight)^{2}}$
$\ \ \ \  = \log{\left( \frac{1 + x}{1 - x} ight) = f(x)}$

Q. If $f(x) = \log\left( \frac{1 + x}{1 - x} \right)$, $- 1 < x < 1$, then $f\left( \frac{3x + x^{3}}{1 + 3x^{2}} \right) - f\left( \frac{2x}{1 + x^{2}} \right)$ is

Explanation

Reviewed by Dr. Abubacker

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