4.5 Partial fractions
Partial fractions run the addition of algebraic fractions backwards: given one complicated fraction, split it into simpler ones. It is needed later for integration.
The form of the split depends on the factors of the denominator.
| Factor in the denominator | Contributes |
| distinct linear, \((x-a)\) | \(\frac {A}{x-a}\) |
| repeated linear, \((x-a)^2\) | \(\frac {A}{x-a}+\frac {B}{(x-a)^2}\) |
| irreducible quadratic, \((x^2+c)\) | \(\frac {Ax+B}{x^2+c}\) |
Solution. Both factors are distinct and linear, so write \[\frac {5x-4}{(x-2)(x+1)}=\frac {A}{x-2}+\frac {B}{x+1}.\] Multiply through by \((x-2)(x+1)\): \[5x-4=A(x+1)+B(x-2).\]
This must hold for every \(x\), so we may choose convenient values.
Put \(x=2\), which kills the \(B\) term: \[5(2)-4=A(3)\quad \implies \quad 6=3A\quad \implies \quad A=2.\]
Put \(x=-1\), which kills the \(A\) term: \[5(-1)-4=B(-3)\quad \implies \quad -9=-3B\quad \implies \quad B=3.\]
\[\therefore \quad \frac {5x-4}{(x-2)(x+1)}=\frac {2}{x-2}+\frac {3}{x+1}.\]
Check. Recombining: \(\frac {2(x+1)+3(x-2)}{(x-2)(x+1)}=\frac {2x+2+3x-6}{(x-2)(x+1)}=\frac {5x-4}{(x-2)(x+1)}\).
Solution. A repeated factor needs both powers: \[\frac {3x+1}{(x-1)^2}=\frac {A}{x-1}+\frac {B}{(x-1)^2}.\] Multiplying by \((x-1)^2\): \[3x+1=A(x-1)+B.\] Put \(x=1\): \(\ 3+1=B\), so \(B=4\). Comparing coefficients of \(x\): \(\ 3=A\). \[\therefore \quad \frac {3x+1}{(x-1)^2}=\frac {3}{x-1}+\frac {4}{(x-1)^2}.\]
Why both terms are needed. Writing only \(\frac {B}{(x-1)^2}\) could never produce the \(3x\); writing only \(\frac {A}{x-1}\) could never produce a denominator of \((x-1)^2\). Omitting one of the two is the usual error with repeated factors.
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