Partial Fractions for Rational Integrands

Compare the numerator's and denominator's degrees. If the numerator's degree is at least the denominator's, divide first. Divide leading terms, multiply by the divisor, subtract, and repeat until the remainder has lower degree. For example, (x^2+6x+4)/(x+2)=x+4-4/(x+2). Integrate the polynomial quotient separately. Apply partial fractions to the proper remainder, whose numerator has lower degree than its denominator.

Factor the proper denominator. Give each distinct linear factor a constant numerator, then multiply by the full denominator and equate numerators. Solve by substitution or by matching coefficients of like powers. The resulting polynomial identity holds for all real x, so denominator zeros may be substituted there. They remain excluded from the original fractions. Check the split by recombining over a common denominator.

For a repeated linear factor (ax+b)^n, include a constant numerator over every power from 1 through n in the full template. Add terms for other factors too. Thus x^2(x+3) needs A/x+B/x^2+C/(x+3). Some coefficients may turn out to be zero. For (px+q)/(x+r)^2, clearing denominators gives px+q=A(x+r)+B=Ax+Ar+B. Match x coefficients and constants: A=p and B=q-pr.

For a≠0, ∫A/(ax+b) dx=(A/a) ln|ax+b|+C. For an integer n>1, ∫B/(ax+b)^n dx=B(ax+b)^(1-n)/(a(1-n))+C. A first-power linear denominator gives a logarithm. Higher powers give rational terms. For example, ∫3/(x-5)^2 dx=-3/(x-5)+C. Differentiate to check the sign, power and scale factor. Work on intervals where the denominator is nonzero.

An irreducible quadratic is a quadratic with no real roots. Its numerator must allow a linear expression. For x(x^2+a^2), with a>0, use A/x+(Bx+D)/(x^2+a^2). If a quadratic factor is repeated, put a separate linear numerator over every required power. A constant numerator alone can miss the x term needed in the general template.

For a>0, ∫(Mx+N)/(x^2+a^2) dx=(M/2) ln(x^2+a^2)+(N/a) arctan(x/a)+C. Split the numerator: substitution u=x^2+a^2 handles Mx, while the constant term uses the arctangent formula. Arctan means inverse tangent, measured in radians. For example, (2x+3)/(x^2+9) integrates to ln(x^2+9)+arctan(x/3)+C. A quadratic piece can contribute both functions.

Combine logarithms using their power and quotient rules. For nonzero u and v, 2 ln|u|-ln|v|=ln|u^2/v|. The coefficient becomes an exponent, and subtraction becomes division inside the logarithm. Combining an answer does not restore any input excluded by the original denominator. A sum of logarithms is also a valid answer; combining them is optional.

For a definite integral, first check the interval for denominator zeros. If the integrand is continuous throughout [a,b] and F is an antiderivative there, evaluate F(b)-F(a). Do not apply this ordinary rule across a denominator zero without a separate check of limits. For the fraction 6/((x-2)(x+4)), F(x)=ln|(x-2)/(x+4)|. From 4 to 8, subtract ln(1/4) from ln(1/2) to get ln 2. Keep logs exact, or round if asked.