Question

find the integral 1/x2-a2 with respect to x and hence evaluate integral 1/4x2-9​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int \dfrac{1}{x^{2} - a^{2} } dx$

As we know that,

Prove

$\sf \implies \displaystyle \int \dfrac{1}{x^{2} - a^{2} } dx \ = \int \dfrac{1}{(x - a)(x + a)} dx$

Multiply and divide by 2a, we get.

$\sf \implies \displaystyle \dfrac{1}{2a} \int \bigg(\dfrac{(x + a) - (x - a)}{(x - a)(x + a)} \bigg)dx$

$\sf \implies \displaystyle \dfrac{1}{2a} \int \bigg(\dfrac{(x + a)}{(x - a)(x + a)} \ - \dfrac{(x - a)}{(x - a)(x + a)} \bigg) dx$

$\sf \implies \displaystyle \dfrac{1}{2a} \int \bigg( \dfrac{dx}{(x - a)} \ - \dfrac{dx}{(x + a)} \bigg)$

$\sf \implies \displaystyle \dfrac{1}{2a} \bigg[ log|x - a| \ - log|x + a| \bigg] + C$

$\sf \implies \displaystyle \dfrac{1}{2a} log \bigg(\dfrac{x - a}{x + a} \bigg) + C$

$\sf \implies \displaystyle \int \dfrac{1}{x^{2} - a^{2} } dx \ = \dfrac{1}{2a} log \bigg( \dfrac{x - a}{x + a} \bigg) + C.$

Question.

$\sf \implies \displaystyle \int \dfrac{dx}{4x^{2} - 9}$

$\sf \implies \displaystyle \int \dfrac{dx}{(2x)^{2} - (3)^{2} }$

As we know that,

Formula of :

$\sf \implies \displaystyle \int \dfrac{1}{x^{2} - a^{2} } dx \ = \dfrac{1}{2a} log \bigg( \dfrac{x - a}{x + a} \bigg) + C.$

$\sf \implies \displaystyle \int \dfrac{dx}{(2x - 3)(2x + 3)}$

$\sf \implies \displaystyle \dfrac{1}{12} \int \dfrac{(2x + 3) - (2x - 3)}{(2x - 3)(2x + 3)} dx$

$\sf \implies \displaystyle \dfrac{1}{12} \int \dfrac{(2x + 3)}{(2x - 3)(2x + 3)} dx \ - \dfrac{1}{12} \int \frac{(2x - 3)}{(2x - 3)(2x + 3)}dx$

$\sf \implies \displaystyle \dfrac{1}{12} \int \bigg( \dfrac{dx}{(2x - 3)} \ - \dfrac{dx}{(2x + 3)} \bigg)$

$\sf \implies \displaystyle \dfrac{1}{12} \bigg[log |2x - 3| \ - log|2x + 3| \bigg]+ C$

$\sf \implies \displaystyle \dfrac{1}{12} log \bigg| \dfrac{(2x - 3)}{(2x + 3)} \bigg| + C.$

$\sf \implies \displaystyle \int \dfrac{dx}{4x^{2} - 9} \ = \dfrac{1}{12} log \bigg| \dfrac{(2x - 3)}{(2x + 3)} \bigg| + C$