integration using partial fraction with repeating denominator 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?
 A: 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:
$$\begin{align}&\int\left(-\frac1{x}-\frac1{x^2}-\frac1{x^3}+\frac1{x-1}\right)dx\\=&-\ln x+\frac1{x}+\frac1{2x^2}+\ln(x-1)\end{align}$$
A: 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.
A: 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.
