If $\frac{d}{dx}\left\lbrack a an^{- 1}{x + b\log\left( \frac{x - 1}{x + 1} ight)} ight brack = \frac{1}{x^{4} - 1}$, then $a - 2b$ is equal to

Question

If $\frac{d}{dx}\left\lbrack a	an^{- 1}{x + b\log\left( \frac{x - 1}{x + 1} ight)} ightbrack = \frac{1}{x^{4} - 1}$, then $a - 2b$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given, $\frac{d}{dx}\left\lbrack a	an^{- 1}{x + b\log\left( \frac{x - 1}{x + 1} ight)} ightbrack = \frac{1}{x^{4} - 1}$
On integrating both sides, we get
$a	an^{- 1}{x + b\log{\left( \frac{x - 1}{x + 1} ight) = \frac{1}{2}}}\ \int_{}^{}\left\lbrack \frac{1}{x^{2} - 1} - \frac{1}{x^{2} + 1} ightbrack\ dx$
$\Rightarrow a	an^{- 1}{x + b\log{\left( \frac{x - 1}{x + 1} ight) = \frac{1}{4}}}\ \int_{}^{}{\log{\left( \frac{x - 1}{x + 1} ight) - \frac{1}{2}}}	an^{- 1}x$
$\Rightarrow \ a = - \frac{1}{2},\ \ b = \frac{1}{4}$
$	herefore\ a - 2b = - \frac{1}{2} - 2\left( \frac{1}{4} ight) = - 1$

Q. If $\frac{d}{dx}\left\lbrack a\tan^{- 1}{x + b\log\left( \frac{x - 1}{x + 1} \right)} \right\rbrack = \frac{1}{x^{4} - 1}$, then $a - 2b$ is equal to

Explanation

Reviewed by Dr. Abubacker

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