Question

1) Evaluate the integration of 1/x²-9​

Answer 1

Solution:-

$\implies\int\bf \dfrac{1}{x^2-9}dx$

Simplify the equation

$\implies\bf\int\dfrac{1}{(x^2-3^2)} dx$

Using the identities ( a - b )(a + b) = a² - b²

$\bf\implies\int \dfrac{1}{( x - 3 )( x + 3 )} dx$

Now using partial Fraction Method

$\implies \bf \dfrac{1}{(x+3)(x-3)} =\dfrac{A}{(x+3)} + \dfrac{B}{(x-3)}$

$\bf\implies 1= A(x+3)+B(x-3)$

When x = -3

$\bf\implies 1 = A( -3 +3 ) + B ( -3-3)$

$\bf\implies 1= 0 - 6B\\\bf\implies B = \dfrac{-1}{6}$

When x = 3

$\bf\implies 1 = A( 3 +3 ) + B ( 3-3)$

$\bf\implies A= \dfrac{1}{6}$

Now we get

$\implies \bf \dfrac{1}{(x+3)(x-3)} =\dfrac{1}{6(x+3)} - \dfrac{1}{6(x-3)}$

Now

$\implies \bf \int\dfrac{1}{6(x+3)}dx - \int\dfrac{1}{6(x-3)}dx$

$\implies \bf \dfrac{1}{6} \int\dfrac{dx}{(x+3)} - \dfrac{1}{6} \int\dfrac{dx}{(x-3)}$

$\bf\implies\dfrac{1}{6} ln|x+3|-\dfrac{1}{6} ln|x-3|+c$

Answer

$\bf\implies\dfrac{1}{6} ln|x+3|-\dfrac{1}{6} ln|x-3|+c$