7.10 Partial Fraction Decomposition
We won’t be using this technique until Question 15.4.10, so feel free to skip this section, although it would be good practice for many of the skills you’ve learned so far.
We know from Question 7.4.19 that we can rewrite \[\frac{a}{b} + \frac{c}{d}\] as \[\frac{ad + cb}{bd}.\]
Question 7.10.1
Sometimes we want to do the reverse of that. Rewrite \[\frac{2x+1}{(x+3)(2x-1)}\] in the form \[\frac{A}{x+3} + \frac{B}{2x-1}\] where $A$ and $B$ are real numbers. The reason we might want to do this is that the second expression is easier to integrate (Section 15.4: Integration) than the first. Show hint
Let’s work backwards from the answer we want: \begin{align*} \frac{A}{x+3} + \frac{B}{2x-1} &= \frac{A(2x-1) + B(x+3)}{(x+3)(2x-1)}\\ &= \frac{2Ax - A + Bx + 3B}{(x+3)(2x-1)}\\ &= \frac{(2A + B)x + (3B-A)}{(x+3)(2x-1)}. \end{align*} We need to find values for $A$ and $B$ so that the numerator is $2x+1$. So we want $2A+B = 2$ and $3B-A = 1$. Solving these simultaneous equations gives $A = 5/7$ and $B = 4/7$.
Question 7.10.2
Rewrite \[\frac{7x-13}{x^2 - 5x + 4}\] in the form \[\frac{A}{Cx+D} + \frac{B}{Ex+F}.\] Show hint
If we can factorise the denominator, then this will become the same sort of question as the previous one we just solved. The denominator, $x^2 - 5x + 4$, factorises to $(x-1)(x-4)$. \begin{align*} \frac{A}{x-1} + \frac{B}{x-4} &= \frac{A(x-4) + B(x-1)}{(x-1)(x-4)}\\ &= \frac{Ax - 4A + Bx - B}{x^2 - 5x + 4}\\ &= \frac{x(A + B) + (-4A - B)}{x^2 - 5x + 4} \end{align*} We want the numerator to match $7x - 13$, so by equating coefficients we get two equations: \begin{align*} A + B &= 7\\ -4A - B &= -13 \end{align*} which have the solution $A = 2$, $B = 5$. Therefore, the original expression can be written as: \[\frac{2}{x-1} + \frac{5}{x-4}.\]
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