How to integrate using partial fractions

Partial fraction decomposition rewrites a complex rational expression as a sum of simpler fractions that can be evaluated using standard integration rules. The method applies when integrating a proper rational function P(x)/Q(x)P(x)/Q(x) where the denominator Q(x)Q(x) can be factored into linear or irreducible quadratic polynomials.

The setup

Check the degrees of the numerator P(x)P(x) and denominator Q(x)Q(x). If the degree of P(x)P(x) is greater than or equal to the degree of Q(x)Q(x), perform polynomial long division first to express the integrand as a polynomial plus a proper rational function. Then, factor the denominator Q(x)Q(x) completely into linear factors (ax+b)(ax+b) and irreducible quadratic factors (ax2+bx+c)(ax^2+bx+c).

The steps

  1. Write the partial fraction expansion based on the factors of Q(x)Q(x). For every non-repeated linear factor (ax+b)(ax+b), add a term A/(ax+b)A/(ax+b). For every irreducible quadratic (ax2+bx+c)(ax^2+bx+c), add a term (Ax+B)/(ax2+bx+c)(Ax+B)/(ax^2+bx+c). For repeated factors, add terms for each power up to the multiplicity. 2. Multiply both sides of the equation by Q(x)Q(x) to clear the denominators. 3. Solve for the unknown constants by substituting strategic values of xx (roots of the factors) or by expanding and equating the coefficients of like powers of xx. 4. Substitute the constants back into the partial fraction expansion. 5. Integrate each resulting simpler term individually, typically yielding logarithmic or arctangent functions.

Checking the result

To verify the integration, differentiate the final answer. Combine the resulting terms over a common denominator algebraically. The simplified result must exactly match the original integrand P(x)/Q(x)P(x)/Q(x).

Common errors

Failing to perform polynomial long division on improper fractions before attempting decomposition. Incorrectly setting up repeated roots by omitting lower-power fractions. Forgetting the absolute value signs when integrating to yield natural logarithms, e.g., writing ln(x)\\ln(x) instead of lnx\\ln|x|. Omitting the constant of integration +C+ C.

Worked example

Evaluate the integral intfrac3x+1x2+x,dx\\int \\frac{3x+1}{x^2+x} \\, dx

The numerator degree is 1 and the denominator degree is 2, so it is a proper fraction. Factor the denominator: x2+x=x(x+1)x^2+x = x(x+1). Set up the decomposition: frac3x+1x(x+1)=fracAx+fracBx+1\\frac{3x+1}{x(x+1)} = \\frac{A}{x} + \\frac{B}{x+1}. Multiply by x(x+1)x(x+1) to clear denominators: 3x+1=A(x+1)+Bx3x+1 = A(x+1) + Bx. Solve for constants: Let x=0x = 0, then 1=A(1)+B(0)1 = A(1) + B(0), so A=1A = 1. Let x=1x = -1, then 2=A(0)+B(1)-2 = A(0) + B(-1), so B=2B = 2. Rewrite the integral: intleft(frac1x+frac2x+1right),dx\\int \\left( \\frac{1}{x} + \\frac{2}{x+1} \\right) \\, dx. Integrate each term: lnx+2lnx+1+C\\ln|x| + 2\\ln|x+1| + C

FAQ

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References: Stewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Chapter 7.4. · OpenStax (2020). Calculus Volume 2. Section 3.4: Partial Fractions.

See also