# Converting into partial fraction.

For the equation,

x-1/(x+1)(x-2)^2 = A/(x+1) + B/(x-2) + C/(x-2)^2

            x-1 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)


Using x = -1 gives A = -2/9

Using x = 2 gives C = 1/3

I am however unable to find the value of B

• Consider $x=0$ and use the values of A and C. – mfl Sep 28 '16 at 12:58
• Or use $x = 1$. Alternatively, since it shall hold for all $x$, compare the coefficients of $x^2$ on both sides. – Daniel Fischer Sep 28 '16 at 12:59
• Or insert any value different from $-1$ and $2$ into $x$ – Peter Sep 28 '16 at 13:02
• you should look at a simple proof, the general algorithm is then obvious – reuns Sep 28 '16 at 13:55

$$\frac { x-1 }{ \left( x+1 \right) { \left( x+2 \right) }^{ 2 } } =\frac { A }{ x+1 } +\frac { B }{ x+2 } +\frac { C }{ { \left( x+2 \right) }^{ 2 } } \\ x-1=A{ \left( x+2 \right) }^{ 2 }+B\left( x+1 \right) \left( x+2 \right) +C\left( x+1 \right) \\ x-1=\left( A+B \right) { x }^{ 2 }+\left( 4A+3B+C \right) x+4A+2B+C\\ \\ \\ \quad \begin{cases} A+B=0 \\ 4A+3B+C=1 \\ 4A+2B+C=-1 \end{cases}\Rightarrow \begin{cases} A=-B \\ -4B+3B+C=1 \\ -4B+2B+C=-1 \end{cases}\Rightarrow \begin{cases} -B+C=1 \\ -2B+C=-1 \end{cases}\Rightarrow \begin{cases} B=2 \\ A=-2 \\ C=3 \end{cases}\\ \\ \\$$
As an alternative to solving a linear system, you may compute a couple of limits, since by assuming $$f(x)=\frac{x-1}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2} \tag{1}$$ we have: $$A=\lim_{x\to -1}(x+1)\,f(x) = \lim_{x\to -1}\frac{x-1}{(x+2)^2}=-2\tag{2}$$ $$C=\lim_{x\to -2}(x+2)^2\,f(x) = \lim_{x\to -2}\frac{x-1}{x+1}=3\tag{3}$$ and $$f(x)-\frac{A}{x+1}-\frac{C}{(x+2)^2} = \frac{2}{x+2} \tag{4}$$ (i.e. $B=2$) follows by direct computation.