I'm supposed to evaluate the following integral by decomposing the integrand into a sum of partial fractions. Note, this question isn't about evaluating.
$$\int\frac{6x+6}{(x^2+1)(x-1)^3}dx$$
From my understanding the numerator should be equated to the following form:
$$\frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}+\frac{E}{(x-1)^3}$$
I know how to evaluate for constants $A$ through $E$, but I'm still unsure as to how the above form was equated to the integrand. If someone could help clarify that'd be great.
By setting $f(x)=\frac{6x+6}{(x^2+1)(x-1)^3}$ we have: $$ E = \lim_{x\to 1}(x-1)^3 f(x) = \lim_{x\to 1}\frac{6x+6}{x^2+1} = 6 \tag{E} $$ $$ D = \lim_{x\to 1}\frac{d}{dx}(x-1)^3 f(x) = -\lim_{x\to 1}\frac{6(x^2+2x-1)}{(x^2+1)^2}=-3\tag{D} $$ $$ C = \frac{1}{2}\lim_{x\to 1}\frac{d^2}{dx^2}(x-1)^3 f(x) = 6\lim_{x\to 1}\frac{(x-1)(1+4x+x^2)}{(1+x^2)^3}=0 \tag{C} $$ and $$ f(x)-\frac{C}{x-1}-\frac{D}{(x-1)^2}-\frac{E}{(x-1)^3} = \frac{3}{1+x^2}\tag{A,B}$$ so $(A,B,C,D,E)=\color{red}{(0,3,0,-3,6)}$. This method clearly shows the strong relation between partial fraction decomposition and the residue theorem. The reason for such a decomposition to hold is that $f(x)$ is a meromorphic function with a triple pole at $x=1$ and simple poles at $x=\pm i$. If we remove from $f(x)$ the "polar part" given by the triple pole at $x=1$, i.e. if we consider $$g(x)=f(x)-\frac{C}{x-1}-\frac{D}{(x-1)^2}-\frac{E}{(x-1)^3}$$ we know in advance that such a meromorphic function is regular in a neighbourhood of $x=1$ and has two simple poles at $x=\pm i$. By removing from $g(x)$ the "polar part" associated with such poles, i.e. by subtracting from $g(x)$ something of the form $\frac{Ax+B}{x^2+1}$, we get that $$p(x)=f(x)-\frac{Ax+B}{x^2+1}-\frac{C}{x-1}-\frac{D}{(x-1)^2}-\frac{E}{(x-1)^3}$$ is an entire function and a rational function, i.e. a polynomial. By inspecting the removed part and the degrees of the polynomials whose ratio defines $f(x)$, we may also easily get what the degree of $p(x)$ has to be, and deduce $$f(x)=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}+\frac{E}{(x-1)^3}.$$ $p(x)\equiv 0$ also follows from computing $\lim_{x\to +\infty}f(x)$ and $\lim_{x\to +\infty}$ of the removed part.