Partial fraction expansion inquiry

Question

How can I expand $\frac{a + 5}{(a^2-1)(a+2)}$ so that the sum of partial fractions is equal to $\frac{1}{a-1} - \frac{2}{a+1} + \frac{1}{a+2}$ ?

Thanks in advance!

Answer 1

We consider the equation $$\frac{a+5}{(a-1)(a+1)(a+2)} = \frac{x}{a-1}+\frac{y}{a+1}+\frac{z}{a+2}$$ and get $$a+5 = (a+1)(a+2)x + (a-1)(a+2)y + (a-1)(a+1)z.$$ Letting $a = 1$ yields $6 = 6x$, so $x = 1$. Now letting $a = -1$, we get $y = -2$ and letting $a = -2$ we get $z = 1$. Thus we know the coefficients.

Answer 2

$$\frac{a + 5}{(a^2-1)(a+2)}=\frac{A}{a-1} + \frac{B}{a+1} + \frac{C}{a+2}$$

$$A(a+1)(a+2)+B(a-1)(a+2)+C(a-1)(a+1) = a+5$$

$$a=1 \implies A=1$$

$$ a=-1 \implies B=-2 $$

$$ a=-2 \implies C=1 $$

Answer 3

$$\frac{a+5}{(a^2-1)(a+2)}=\frac{a+5}{(a-1)(a+1)(a+2)}=\frac{a-1+6}{a-1}\left(\frac{1}{a+1}-\frac{1}{a+2}\right)=$$ $$=\frac{1}{a+1}-\frac{1}{a+2}+\frac{6}{(a-1)(a+1)}-\frac{6}{(a-1)(a+2)}=$$ $$=\frac{1}{a+1}-\frac{1}{a+2}+3\left(\frac{1}{a-1}-\frac{1}{a+1}\right)-2\left(\frac{1}{a-1}-\frac{1}{a+2}\right)=$$ $$=\frac{1}{a-1}-\frac{2}{a+1}+\frac{1}{a+2}.$$