Question

Partial fraction of x/(x+1) (x+2)² is
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Answer 1

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

Let assume that

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

On taking LCM, we get

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

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

$\rm \: - 1 = a {( - 1 + 2)}^{2} + b( - 1 + 1)( - 1 + 2) + c( - 1 + 1)$

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

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

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

$\rm \: - 2 = a {( - 2 + 2)}^{2} + b( - 2 + 1)( - 2 + 2) + c( - 2 + 1)$

$\rm \: - 2 = 0 + 0 - c$

$\rm\implies \:\boxed{ \rm{ \:c \: = \: 2 \: \: }}$

On substituting x = 0, in equation (2), we get

$\rm \:0 = a {(0 + 2)}^{2} + b(0 + 1)(0 + 2) + c(0 + 1)$

$\rm \:0 = 4a + 2b + c$

On substituting the values of a and c, we get

$\rm \:0 = - 4 + 2b + 2$

$\rm \:0 = - 2 + 2b$

$\rm \: 2b = 2$

$\rm\implies \:\boxed{ \rm{ \:b \: = \: 1 \: \: }}$

So, on substituting the values of a, b and c in equation (1), we get

$\boxed{ \rm{\dfrac{x}{(x + 1) {(x + 2)}^{2} } = - \dfrac{1}{x + 1} + \dfrac{1}{x + 2} + \dfrac{2}{ {(x + 2)}^{2} }}}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Fraction & \bf Partial \:  Fraction \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \dfrac{1}{(ax + b)(cx + d)}  & \sf \dfrac{p}{ax + b}  + \dfrac{q}{cx + d}  \\ \\ \sf \dfrac{1}{ {(ax + b)}^{2} }  & \sf \dfrac{p}{ax + b}  + \dfrac{q}{ {(ax + b)}^{2} }  \\ \\ \sf \dfrac{1}{(ax + b)(c {x}^{2}  + d)}  & \sf \dfrac{p}{ax + b}  + \dfrac{qx + r}{c {x}^{2}  + d}  \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

Partial Fraction Decomposition begins.

$\\ \tt\frac{x}{(x+1){(x+2)}^{2}}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{{(x+2)}^{2}}$

Rewrite the right side of the equation with a common denominator.

$\\ \tt\frac{x}{(x+1){(x+2)}^{2}}=\frac{A{(x+2)}^{2}+B(x+1)(x+2)+C(x+1)}{(x+1){(x+2)}^{2}}$

Cancel (x+1)(x+2)².

$\\ \tt x=A{(x+2)}^{2}+B(x+1)(x+2)+C(x+1)$

$\\ \tt x=A{x}^{2}+4Ax+4A+B{x}^{2}+3Bx+2B+Cx+C$

Group terms by degrees of x.

x=(A+B)x²+(4A+3B+C)x+4A+2B+C

Set up an equation system based on the degrees of x.

$\\ \begin{aligned}& \tt \: A+B=0\\& \tt4A+3B+C=1\\& \tt4A+2B+C=0\end{aligned}$

Substitute A,B,C into the original expression.

$\\ \tt\frac{x}{(x+1){(x+2)}^{2}}=-\frac{1}{x+1}+\frac{1}{x+2}+\frac{2}{{(x+2)}^{2}}$

Partial Fraction Decomposition completes.

$\\ \tt-\frac{1}{x+1}+\frac{1}{x+2}+\frac{2}{{(x+2)}^{2}}$