<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] From $$
\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}
$$ you have: $$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}=\frac{3x+2}{x(x+1)^3}$$ Now let $x=0$ then $$A(0+1)^3=3\times0+2\to A=2$$ Let $x=-1$ then $$D(-1)=3\times(-1)+2=-1\to D=-1$$ Set $x=1$ then $$8A+4B+2C+D=3\times1+2=5\to 16+4B+2C-1=5\to4B+2C=-10$$
Set $x=2$ then $$27A+18B+6C+2D=3\times 2+2=8\to 54+18B+6C-2=8\to18B+6C=-44$$ Now solve the two final equation respect to $B$ and $C$.
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