Techniques of Integration · Partial Fractions
Partial Fractions: Irreducible Quadratic
For irreducible quadratic factors (discriminant < 0), the numerator is a linear expression Ax + B.
Worked examples
Find ∫ 1/(x(x²+1)) dx.
- 1/(x(x²+1)) = A/x + (Bx+C)/(x²+1)
- 1 = A(x²+1) + (Bx+C)x. x=0: 1=A. Expand: 1=(A+B)x²+Cx+A
- A+B=0 → B=-1. C=0.
- ∫ [1/x + (-x)/(x²+1)] dx = ln|x| - (1/2)ln(x²+1) + C
Answer: ln|x| - (1/2)ln(x²+1) + C
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