Converting a proper fraction into partial fraction

Question

For solving integration-related questions, a rational proper fraction of the form $\frac{px^{2}+qx+r}{(x-a)(x^{2}+bx+c)}$ is decomposed into the sum of the expressions, $$\frac{A}{x-a} + \frac{Bx+C}{x^{2}+bx+c}$$ where $x^{2}+bx+c$ can't be factorised further.

I don't understand why we have to use $Bx+C$ in the numerator of the second expression. Why can't it be just $B$?

To elaborate, if $\frac{x^{2}+x+1}{(x-4)(x^{2}+x+3)}$ is decomposed into the sum $\frac{A}{x-4} + \frac{B}{x^{2}+x+3}$, we will get contradictory values for $A$ and $B$. But if we use $Bx+C$ in the second expression, the method works fine. Why is this so?

Answer 1

In full, the partial fraction decomposition for OP's expression should be

$$\frac{p(x)}{q(x)} = \frac{a_1}{x-r_1} + \frac{a_2}{x-r_2} + \frac{a_3}{x-r_3} \tag{$*$}$$

where $a_i\in\Bbb C$ and $r_i$ denote the roots of $q(x)$. But any two of the fractions on the RHS can be joined to obtain e.g.

$$\begin{align*} \frac{p(x)}{q(x)} &= \frac{a_1}{x-r_1} + \frac{a_2\left(x-r_3\right)+a_3\left(x-r_2\right)}{\left(x-r_2\right)\left(x-r_3\right)} \\ &= \frac{a_1}{x-r_1} + \frac{\left(a_2+a_3\right)x - \left(a_2r_3+a_3r_2\right)}{x^2 - \left(r_2+r_3\right)x + r_2r_3} \tag{$**$} \end{align*}$$

The expansion in $(**)$ is easier to work with in OP's example since $x^2+x+3=0$ at the roots $r_{2,3}=\dfrac{-1\pm i\sqrt{11}}2$. There's far less (read: no) juggling of imaginary constants involved.

By the same token, if $q(x)$ has any repeated factors, we have the "shortcut" e.g.

$$\frac{p(x)}{\left(x-r_1\right) \left(x-r_2\right)^2} = \frac{a_1}{x-r_1} + \frac{a_2}{x-r_2} + \frac{a_3}{\left(x-r_2\right)^2}$$

since

$$\frac{a_2}{x-r_2} + \frac{a_3}{\left(x-r_2\right)^2} = \frac{a_2 x + \left(a_3- a_2 r_2\right)}{x^2 - 2r_2x + r_2^2}$$

Answer 2

This one doesn't fit in comments so

Take $$ \frac{x^2+px+q}{(x+1)(x^2+2x+3)}=\frac{B}{x^{2}+2x+3}+\frac{A}{x+1} $$

Cross multiplying you'd get

$$ x^2+px+q=B(x+1)+A(x^2+2x+3)=Ax^2+(2A+B)x+(3A+B) $$

This gives $$ A=1\\ 2A+B=p\\ 3A+B=q $$

Some cases these equations are consistent mostly they aren't and when they aren't they don't have solutions

To allow for solutions we relax the conditions and add a third variable

Answer 3

The ${A\over x-a}$ term is irrelevant; the question is why ${Bx+C\over x^2+bx+c}$ can't be identically equal to ${D\over x^2+bx+c}$ for some number $D$. But $$ {Bx+C\over x^2+bx+c}={D\over x^2+bx+c} $$ on clearing denominators is just $Bx+C=D$, which can't be true for all $x$ unless $B=0$.