Question

solve(x-1)/ (x+3)(x²-4) into proper fraction​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\dfrac{x - 1}{(x + 3)( {x}^{2} - 4)}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{x - 1}{(x + 3)( x - 2)(x + 2)}$

$\rm\:Let \: \dfrac{x - 1}{(x + 3)( x - 2)(x + 2)} = \dfrac{a}{x + 3} + \dfrac{b}{x - 2} + \dfrac{c}{x + 2} - - (1)$

$\rm \: x - 1 = a(x + 2)(x - 2) + b(x + 3)(x + 2) + c(x + 3)(x - 2) - - (2)$

On substituting x = - 3 in equation (2) we get

$\rm \: - 3 - 1 = a( - 3+ 2)( - 3- 2) + b( - 3+ 3)( - 3 + 2) + c( - 3 + 3)( - 3 - 2)$

$\rm \: - 3 - 1 = a( -1)( - 5)$

$\bf\implies \:a = - \: \dfrac{4}{5} - - - (3)$

On substituting x = 2 in equation (2), we get

$\rm \: 2 - 1 = a(2 + 2)(2 - 2) + b(2 + 3)(2 + 2) + c(2 + 3)(2 - 2)$

$\rm \: 1 = b(5)(4)$

$\bf\implies \:b \: = \: \dfrac{1}{20} - - - (4)$

On substituting x = - 2 in equation (2), we get

$\rm \: - 2 - 1 = a( - 2 + 2)( - 2 - 2) + b( - 2 + 3)( - 2 + 2) + c( - 2 + 3)( - 2 - 2)$

$\rm \: - 3 = c( 1)( - 4)$

$\bf\implies \:c = \dfrac{3}{4} - - - (5)$

On substituting the values of a, b, c in equation (1), we get

$\rm\: \therefore \: \: \dfrac{x - 1}{(x + 3)( x - 2)(x + 2)} = \dfrac{ - 4}{5(x + 3)} + \dfrac{1}{20(x - 2)} + \dfrac{3}{4(x + 2)}$