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Chain Rule of Differentiation

∫1/x2 - 1ln x...

Question

$\int \frac{1}{x^{2} - 1} l n \frac{x - 1}{x + 1} d x$ equals

A

\frac{1}{2} l n^{2} ∣ ∣ ∣ \frac{x - 1}{x + 1} ∣ ∣ ∣ + c

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B

\frac{1}{4} l n^{2} ∣ ∣ ∣ \frac{x - 1}{x + 1} ∣ ∣ ∣ + c

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C

\frac{1}{2} l n^{2} ∣ ∣ ∣ \frac{x + 1}{x - 1} ∣ ∣ ∣ + c

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D

\frac{1}{4} l n^{2} ∣ ∣ ∣ \frac{x + 1}{x - 1} ∣ ∣ ∣ + c

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Solution

The correct options are
B $\frac{1}{4} l n^{2} ∣ ∣ ∣ \frac{x - 1}{x + 1} ∣ ∣ ∣ + c$
D $\frac{1}{4} l n^{2} ∣ ∣ ∣ \frac{x + 1}{x - 1} ∣ ∣ ∣ + c$
$I = \int \frac{1}{x^{2} - 1} l n ∣ ∣ ∣ \frac{x - 1}{x + 1} ∣ ∣ ∣ d x$
Put $l n ∣ ∣ ∣ \frac{x - 1}{x + 1} ∣ ∣ ∣ = t \Rightarrow \frac{1}{2 (x^{2} - 1)} d x = d t$
$I = \frac{1}{2} \int t d t = \frac{1}{4} t^{2}$

$= \frac{1}{4} {(l n ∣ ∣ ∣ \frac{x - 1}{x + 1} ∣ ∣ ∣)}^{2}$

$= \frac{1}{4} {(- l n ∣ ∣ ∣ \frac{x + 1}{x - 1} ∣ ∣ ∣)}^{2}$

$= \frac{1}{4} {(l n ∣ ∣ ∣ \frac{x + 1}{x - 1} ∣ ∣ ∣)}^{2}$

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