Question

Evaluate the Integral :-

${ \bigstar { \underline { \boxed { \displaystyle { \sf { \int { \dfrac{1}{x² - 1} dx } } } } } } { \bigstar} }$ ​

Answer 1

Step-by-step explanation:

$thank \: you$

Answer 2

⇝ To Find :-

Value of $\displaystyle { \rm { \int { \dfrac{1}{x^2 - 1} dx }}}$

⇝ Solution :-

❒ Method 1 :

As

$\rm \frac{1}{ {x}^{2} - 1}$

$= \rm \frac{1}{ {x}^{2} - 1 {}^{2} }$

$= \rm \frac{1}{(x + 1)(x - 1)}$

$= \rm \frac{1}{2} \bigg( \frac{1}{x - 1} - \frac{1}{x + 1} \bigg)$

Now,

$\therefore\displaystyle { \rm { \int { \dfrac{1}{x^2 - 1} dx }}}$

$= \displaystyle { \rm { \int \frac{1}{2} \bigg( { \dfrac{1}{x - 1} - \frac{1}{x + 1} \bigg)dx }}}$

$= \frac{1}{2} \displaystyle { \rm { \int \bigg( { \dfrac{1}{x - 1} - \frac{1}{x + 1} \bigg)dx }}}$

$= \frac{1}{2} \bigg[\displaystyle { \rm { \int \bigg( { \dfrac{1}{x - 1} \bigg)dx }}} - { \int \rm\bigg( { \frac{1}{x + 1} \bigg)dx }} \bigg]$

$= \rm \frac{1}{2} \bigg( \log |x - 1| - \log |x + 1| \bigg) + C$

$\purple{ \rm= \frac{1}{2} \bigg( \log \bigg| \frac{x - 1}{x + 1} \bigg| \bigg)+C }$

❒ Method 2 :-

$\rm\therefore\int \frac{1}{ {x}^{2} - 1 }dx$

$\rm = \int \frac{1}{ {x}^{2} - {1}^{2} } dx$

We Have Formula that

$\bf \red\bigstar \: \: \orange{ \underbrace{ \underline{  \int \frac{1}{ {x}^{2} - {a}^{2} } \: dx = \frac{1}{2a} log \bigg| \frac{x - a}{x + a} \bigg| +C }}}$

$\rm Here \: \: \: a = 1$

Therefore,

$\rm \int \frac{1}{ {x}^{2} - {1}^{2} } dx$

$\purple{ \rm= \frac{1}{2} \bigg( \log \bigg| \frac{x - 1}{x + 1} \bigg| \bigg)+C }$