Question

resolve into partial fractions​

Answer 1

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

Given that

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

To resolve in to partial fraction,

Let assume that

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

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

On substituting 'x = - 1', we get

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

$\rm :\longmapsto\: - 2 + 3 = - 4a$

$\rm :\longmapsto\: 1 = - 4a$

$\rm :\implies\: \boxed{ \bf{ \: a = - \: \frac{1}{4}}}$

On substituting 'x = 3', we get

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

$\rm :\longmapsto\:6 + 3 = 4b$

$\rm :\longmapsto\:9 = 4b$

$\rm :\implies\: \boxed{ \bf{ \: b = \: \frac{9}{4}}}$

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

$\bf :\longmapsto\:\dfrac{2x + 3}{(x + 1)(x - 3)} = \dfrac{ - 1}{4(x + 1)} + \dfrac{9}{4(x - 3)}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf form \: of \: rational \: {f}^{n} & \bf form \: of \: partial \: fraction \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \dfrac{1}{(ax + b)(cx + d)} & \sf \dfrac{p}{ax + b} + \dfrac{q}{cx + d} \\ \\ \sf \dfrac{1}{ {(ax + b)}^{2} } & \sf \displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} \\ \\ \sf \dfrac{1}{(x - d)( {ax}^{2} + bx + c)} & \sf \displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}} + \dfrac{C}{x - d} \end{array}} \\ \end{gathered}$