An elementary question on integrating

Question

My question contains the following example : Find the integral of $$\frac{x^3-x+4}{x^4-2x^3+2x^2-2x+1} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C x + D}{x^2+1}$$

So why is $B$ not multiplied by $x$ with another variable say $E$ because the denominator is also from a 2nd degree?

Answer 1

Notice that we have $$\frac{A}{x-1} + \frac{B}{(x-1)^2}$$ which is indeed $$\frac{Ax-A+B}{(x-1)^2}$$ which also can be written as $$\frac{Ex+F}{(x-1)^2}$$ by defining $E = A$ and $F = -A+B$.

Or we can change some of the definitions to get: $$\frac{A}{x-1} + \frac{Gx+H}{(x-1)^2} = \frac{(A+G)x+H-A}{(x-1)^2} = \frac{Ex+F}{(x-1)^2}$$ where we can define $E = A+G$ and $F = H-A$. These two should give same $E$ and $F$ in the end. But in this case, values for $A,G,H$ are not unique so I suggest you to use the first decomposition.

Answer 2

Because the theorem on partial fraction decomposition of rational functions claims that such a decomposition exists. So, why bother to introduce a new constant $E$? Computations are simpler without it.

Answer 3

The general partial fractions decomposition theorem asserts that`

Let $K$ be a commutative field, $\dfrac{F(x)}{G(x)}$ a rational fraction with coefficients in $K$. This fraction has a unique decomposition as a sum of a polynomial $Q(X)$ and rational fractions $\dfrac{A(x)}{P(x)^n}$, where

• $Q(x)$ is the remainder in the euclidean division of $F(X)$ by $G(X)$,
• $P(X)$ is irreducible in $K[X]$ and $\;\deg A <\deg \color{red}{P}$.

Furthermore, assume the denominator has a decomposition into irreducible polynomials: $\;G(X)=P_1(X)^{n_1}\dotsm P_t(X)^{n_t}.$ Then $$\frac{F(x)}{G(x)}=Q(X)+R_1(X)+\dots+R_t(X),\qquad R_i(X)\in K(X).$$ where each rational fraction has the form $$R_i(X)=\frac{A_{i,1}(X)}{P_{i}(X)}+\dots+\frac{A_{i,n_{i}}(X)}{P_{i}(X)^{n_i}},\qquad \deg A_{i,1},\dots,\deg A_{i,{n_i}}<\deg P_i. $$