<s>[INST] Rational function partial fraction help? So I have to integrate 
$$
\frac{3x+2}{(x)(x+1)^3}
$$ so I have the terms 
$$
\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}
$$ 
... and then I have 
$$
3x+2=A(x+1)^3 + B(x)(x+1)^2 +C(x)(x+1)^2+ D(x)(x+1)^2
$$
... here I tried to equal the quotients and I have $A=2$..how do I find the others? 
 [/INST] You’re starting off on the wrong foot. When you put
$$\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}\tag{1}$$
back together into a single fraction, that fraction must have the same denominator as the original one. $(1)$ becomes
$$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}\;.$$
and you want to choose $A,B,C$, and $D$ so that 
$$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}=\frac{3x+2}{x(x+1)^3}\;.$$
Two fractions with the same denominator are equal if and only their numerators are equal, so you require that
$$A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx=3x+2\;.\tag{2}$$
Setting $x=0$, you find that $A=2$. Setting $x=-1$, you find that $-D=-1$, so $D=1$. Thus, $(2)$ reduces to
$$2(x+1)^3+Bx(x+1)^2+Cx(x+1)+x=3x+2\;,$$ or
$$2(x+1)^3+Bx(x+1)^2+Cx(x+1)=2x+2\;.$$ Now just multiply out the lefthand side:
$$2x^3+6x^2+6x+2+Bx^3+2Bx^2+Bx+Cx^2+Cx=2x+2\;,$$ or
$$(B+2)x^3+(2B+C+6)x^2+(B+C+4)x=0\;.$$
The coefficients on the righthand side are all $0$, so all that remains is to solve
$$\left\{\begin{align*}
&B+2=0\\
&2B+C+6=0\\
&B+C+4=0
\end{align*}\right.$$
for $B$ and $C$. The first equation makes this easy: $B=-2$, and then $C=-2$ as well.
</s>