How to compute the partial fraction decomposition of $\frac{6}{x^4(x+1)}$?

Question

How do I compute the partial fraction decomposition of $\frac{6}{x^4(x+1)}$?

I let $$\frac{6}{x^4(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x^4} + \frac{F}{x+1}$$

When I let $x=-1$, $F=6$ and when $x=0$, $D=6$. How can I find the constants $A, B,$ and $C?$.

Thanks.

Answer 1

One way is to write $$6 = (x+1)x^3 A + (x+1)x^2 B + (x+1)x C + (x+1)D + x^4 F$$ and then after substituting the values you already know, we get $$6 = (x+1)x^3 A + (x+1)x^2 B + (x+1)x C + 6(x+1) + 6x^4.$$ Then collect like terms in $x$: $$6 = (6+A)x^4 + (A + B)x^3 + (B + C)x^2 + (C+6)x + 6.$$ Since this equation must be true for all $x$, this requires all coefficients but the constant term be equal to zero; i.e., $$6+A = 0, \quad A+B = 0, \quad B+C = 0, \quad C+6 = 0,$$ from which it follows that $$A = -6, \quad B = 6, \quad C = -6.$$

Answer 2

You can also do it like this: $$\frac{1}{x(x+1)}=\frac1x-\frac{1}{x+1}$$ Multiply by $1/x$: $$\frac{1}{x^2(x+1)}=\frac1{x^2}-\frac{1}{x(x+1)}=\frac1{x^2}-\frac1x+\frac1{x+1}$$ and again $$\frac{1}{x^3(x+1)}=\frac1{x^3}-\frac1{x^2}+\frac1{x(x+1)}=\frac{1}{x^3}-\frac{1}{x^2}+\frac1x-\frac{1}{x+1}$$ See the pattern and you can prove $$\boxed{\frac{1}{x^n(x+1)}=\frac{1}{x^n}-\frac{1}{x^{n-1}}+\cdots-(-1)^n\frac1x+(-1)^n\frac{1}{x+1}}$$ Set $n=4$ and get $$\frac1{x^4(x+1)}=\frac1{x^4}-\frac1{x^3}+\frac1{x^2}-\frac1x+\frac1{x+1}$$ Hope this helps. :)

Answer 3

Starting from where you left off, we have that

$$\frac 6{x^4(x+1)}=\frac Ax+\frac B{x^2}+\frac C{x^3}+\frac 6{x^4}+\frac 6{x+1}$$

Multiply both sides by $x$ and take the limit as $x\to+\infty$.

$$\lim\limits_{x\to+\infty}\frac 6{x^3(x+1)}=A+6\lim\limits_{x\to+\infty}\frac x{x+1}$$

Solve the coefficients as

$$A=\lim\limits_{x\to+\infty}\left[\frac 6{x^3(x+1)}-\frac Bx-\frac C{x^2}-\frac 6{x^3}-\frac {6x}{x+1}\right]=-6$$

For $B$, we have

$$\begin{align*}B & =\lim\limits_{x\to+\infty}\left[\frac 6{x^2(x+1)}+6x-\frac Cx-\frac 6{x^2}-\frac {6x^2}{x+1}\right]\\ & =\lim\limits_{x\to+\infty}\frac {6(x^2-x+1)}{x^2}\\ & =6\end{align*}$$

And for $C$, we have

$$\begin{align*}C & =\lim\limits_{x\to+\infty}\left[\frac 6{x(x+1)}+6x^2-6x-\frac 6x-\frac {6x^3}{x+1}\right]\\ & =\lim\limits_{x\to+\infty}\frac {6(1-x)}x\\ & =-6\end{align*}$$

This gives

$$\frac 6{x^4(x+1)}=-\frac 6x+\frac 6{x^2}-\frac 6{x^3}+\frac 6{x^4}+\frac 6{x+1}$$