Partial fractions
\(\int\frac{x+1}{(x-1)(x-3)}dx\).
\(\frac{x+1}{(x-1)(x-3)}=\frac{-1}{x-1}+\frac{2}{x-3}\). Integral \(=-\log|x-1|+2\log|x-3|+C=\log\frac{(x-3)^2}{|x-1|}+C\).
Integrals
Integration by Partial Fractions
Understand
A proper rational function \(\frac{P(x)}{Q(x)}\) (degree of \(P\) < degree of \(Q\)) can be split into simpler fractions whose integrals are logarithms or inverse tangents.
| Factor of Q(x) | Partial fraction(s) |
|---|---|
| \((x-a)\) | \(\frac{A}{x-a}\) |
| \((x-a)^2\) | \(\frac{A}{x-a}+\frac{B}{(x-a)^2}\) |
| \(x^2+bx+c\) (irreducible) | \(\frac{Ax+B}{x^2+bx+c}\) |
Example: \(\frac{1}{(x-1)(x+2)}=\frac{1/3}{x-1}-\frac{1/3}{x+2}\), so \(\int\frac{dx}{(x-1)(x+2)}=\frac13\log\left|\frac{x-1}{x+2}\right|+C\).
If the fraction is improper, divide first. A quick way to find constants for linear factors is the "cover-up" rule: \(A=\left.\frac{1}{x+2}\right|_{x=1}=\frac13\).
Key Concepts
- Make the fraction proper (long division).
- Write the correct form for each factor.
- Find constants by substitution or comparing coefficients.
Solved Examples
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Factorise the denominator, split, integrate each simple fraction.
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