Question

Resolve into partial faction
to (X+3)/(X+1) (X-1)​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Resolve \: in \: to \: partial \: fraction : \: \dfrac{x + 3}{(x + 1)(x - 1)}$

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

$\bf \: Let \: \dfrac{x + 3}{(x + 1)(x - 1)} = \sf \: \dfrac{a}{x + 1} + \dfrac{b}{x - 1} - - (1)$

On taking LCM, we get

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

On substituting 'x = - 1' in equation (2), we get

$\rm :\longmapsto\: - 1 + 3 = a( - 1 - 1) + 0$

$\rm :\longmapsto\:2 = - 2a$

$\bf\implies \:a \: = \: - \: 1 - - (3)$

On substituting 'x = 1' in equation (2), we get

$\rm :\longmapsto\:1 + 3 = 0 + b(1 + 1)$

$\rm :\longmapsto\:4 = 2b$

$\bf\implies \:b \: = \: 2 - - (4)$

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

$\bf \: \dfrac{x + 3}{(x + 1)(x - 1)} = \sf \: \dfrac{ - 1}{x + 1} + \dfrac{2}{x - 1}$

Additional Information :-

Partial Fraction :-

Partial fraction decomposition is the breaking down of a rational expression into simplier parts. It is the opposite of adding rational expressions. When adding two rational expressions, there has to be a common denominator.

$\begin{gathered}\boxed{\begin{array}{c|c} \sf term \: in \: denominator & \sf partial \: fraction \: decomposition \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ax + b & \sf \displaystyle \frac{A}{{ax + b}} \\ \\ \sf {(ax + b)}^{2} & \sf \displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} \\ \\ \sf {ax}^{2} + bx + c & \sf \displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}} \end{array}} \\ \end{gathered}$