Integration by Partial Fractions
The idea
When the denominator of a rational function factors into distinct linear pieces, the whole fraction can be rewritten as a sum of simpler fractions, each of which has a known antiderivative involving a logarithm:
∫x−a1dx=ln∣x−a∣+CWorked example
Find ∫(x+1)(x+2)xdx.
Solution: Write (x+1)(x+2)x=x+1A+x+2B, so x=A(x+2)+B(x+1). Setting x=−1 gives A=−1; setting x=−2 gives B=2:
∫(x+1−1+x+22)dx=−ln∣x+1∣+2ln∣x+2∣+C=ln∣x+1∣(x+2)2+CWorked example
Find ∫x2+3x+22xdx.
Solution: The denominator factors as (x+1)(x+2). Writing (x+1)(x+2)2x=x+1A+x+2B gives 2x=A(x+2)+B(x+1). Setting x=−1: A=−2. Setting x=−2: B=4:
∫(x+1−2+x+24)dx=4ln∣x+2∣−2ln∣x+1∣+CContinue learning
Partial fractions are one technique for a rational integrand: some standard rational forms have their own ready-made formulas instead.