integration using partial fraction with repeating denominator

Question

So i need to integrate this $1/[(x^3)(x-1)]$, that means it could be decomposed into:

$$A/x + B/x^2 + C/x^3 + D(x-1)$$

Furthermore the resulting equation would then be:

• $1 = A(x^2)(x-1) + Bx(x-1) + C(x-1) + Dx^3$

• if i let $x=0$ then $C=-1$

• if i let $x=1$ then $D=1$

But i don't know how to get the values for A and B can somebody help me?

Answer 1

Here's an easy way to do it: $$\frac{1}{x^3(x-1)}=\frac{Ax^2+Bx+C}{x^3}+\frac{D}{x-1}$$

I chose to start at $x^2$ for the first one since the degree of the denominator is $3$, and on the other $x^0$ because the denominator has degree $1$ (so go all the way down starting with $x^{\text{denominator degree}-1}$).

Combine the two fractions into a fraction with the original denominator: $$\frac{(Ax^2+Bx+C)(x-1)+D(x^3)}{x^3(x-1)}$$

And recall that the numerator was equal to $1$: $$(Ax^2+Bx+C)(x-1)+D(x^3)=1$$ Distribute: $$Ax^3+Bx^2+Cx-Ax^2-Bx-C+Dx^3=1$$ Combine like terms: $$(A+D)x^3+(B-A)x^2+(C-B)x-C=1$$

And guess what..? $$(A+D)x^3+(B-A)x^2+(C-B)x-C=0x^3+0x^3+0x+1$$

Hence we have: $$\begin{cases}A+D=0\\B-A=0\\C-B=0\\-C=1\end{cases}$$

Which gives you: $$\begin{cases}A=-1\\B=-1\\C=-1\\D=1\end{cases}$$

And so you have: $$\frac{-x^2-x-1}{x^3}+\frac1{x-1}=-\frac1{x}-\frac1{x^2}-\frac1{x^3}+\frac1{x-1}$$

Which is easy to integrate: $$\begin{align}&\int\left(-\frac1{x}-\frac1{x^2}-\frac1{x^3}+\frac1{x-1}\right)dx\\=&-\ln x+\frac1{x}+\frac1{2x^2}+\ln(x-1)\end{align}$$

Answer 2

Hint: Try two more values of $x$ say $x=2$ and $x=-1$, and use your values of $C$ and $D$ to get the values for $A$ and $B$ from the resulting equations. You will have two equations in $2$ unknowns.

Answer 3

If $x=2$ then $1=4A+2B+C+8D$. If $x=-1$ then $1=-2A+2B-2C-D$. Since you know $C,D$ you now have two linear equations in $A,B$ which you can solve.