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Techniques of Integration · Partial Fractions

Partial Fractions: Irreducible Quadratic

P(x)(x2+bx+c)=Ax+Bx2+bx+c\frac{P(x)}{(x^2+bx+c)} = \frac{Ax+B}{x^2+bx+c}

For irreducible quadratic factors (discriminant < 0), the numerator is a linear expression Ax + B.

  • 2 worked examples
  • 16 in Techniques of Integration

Worked examples

Find ∫ 1/(x(x²+1)) dx.
  1. 1/(x(x²+1)) = A/x + (Bx+C)/(x²+1)
  2. 1 = A(x²+1) + (Bx+C)x. x=0: 1=A. Expand: 1=(A+B)x²+Cx+A
  3. A+B=0 → B=-1. C=0.
  4. ∫ [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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Find ∫ (2x + 3)/(x² + 4) dx.

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Common questions

What is the Partial Fractions: Irreducible Quadratic?

For irreducible quadratic factors (discriminant < 0), the numerator is a linear expression Ax + B. It is one of the techniques of integration formulas in the CalcRef reference.

How do you use the Partial Fractions: Irreducible Quadratic?

Worked example. 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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