Question

Evaluate improper integral
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Answer 1

Step-by-step explanation:

We have,

$2 \int \limits^{ - 2} _{ - \infty } \frac{dx}{ {x}^{2} - 1 }$

$=2 \int \limits^{ - 2} _{ - \infty } \frac{dx}{( x- 1 )(x+1)}$

$=4\int \limits^{ - 2} _{ - \infty } \frac{((x+1) - (x-1))dx}{ (x-1)(x+1) }$

$=2 \int \limits^{ - 2} _{ - \infty } \frac{dx}{ x- 1 }-2 \int \limits^{ - 2} _{ - \infty } \frac{dx}{ x +1 }$

$=[ln(x-1)]^{ - 2} _{ - \infty }-[ln(x+1)]^{ - 2} _{ - \infty }$

$= [ln(\frac{x-1}{x+1})]^{ - 2} _{ - \infty }$

$=[ln(\frac{1-\frac{1}{x}}{1+\frac{1}{x}})]^{ - 2} _{ - \infty }$

$=ln(3) -ln(1)$

$= ln(3)$