Read through concept notes with worked explanations and diagrams, chapter by chapter.
Tip: use←→to switch concepts
Concept 04 / 10Integral Calculus
Integration of Rational Functions by Partial Fractions
Reducing improper fractions by division, decomposing proper fractions with distinct linear, repeated linear and irreducible quadratic factors, the cover-up method for constants, and special substitutions such as x^2 = t for even rational functions.
A rational function Q(x)P(x) is proper if degP<degQ. If it is improper, first divide: QP=S(x)+Q(x)R(x) with degR<degQ. The polynomial S integrates directly; the proper part is split into partial fractions, each of which integrates to a logarithm, a power, or an inverse tangent.
Decomposition rules for the proper fraction Q(x)R(x), after factoring Q completely over the reals:
Distinct linear factor (x−a): a term x−aA.
Repeated linear factor (x−a)k: terms x−aA1+(x−a)2A2+⋯+(x−a)kAk. All powers up to k are needed.
Irreducible quadratic factor (x2+px+q) with p2<4q: a term x2+px+qBx+C, with a linear numerator.
Repeated irreducible quadratics similarly need one term for each power.
Finding the constants. Multiply through by Q(x) to clear denominators, then either substitute convenient values of x (the roots of the linear factors kill all but one term) or compare coefficients. For distinct linear factors, the cover-up method is fastest: the constant A for (x−a) equals Q1(a)R(a), where Q1 is Q with the factor (x−a) covered up. For example, (x−1)(x+2)2x+3 has A=35 at x=1 and B=−3−1=31 at x=−2. For a repeated factor, cover-up gives only the highest-power coefficient; find the rest by substituting another value or comparing coefficients.
Integrating the pieces.
∫x−aAdx=Aln∣x−a∣.
∫(x−a)kAdx=(k−1)(x−a)k−1−A for k≥2.
∫x2+px+qBx+Cdx: write the numerator as 2B(2x+p)+(C−2Bp). The first part gives 2Bln(x2+px+q), and the second, after completing the square, gives an inverse tangent.
Useful standard results that come from partial fractions:
∫x2−a2dx=2a1lnx+ax−a+C and ∫a2−x2dx=2a1lna−xa+x+C.
Shortcuts for special structures. If the integrand is a function of x2 only (like (x2+1)(x2+4)x2), treat x2 as a single variable t when decomposing (not when integrating): (t+1)(t+4)t=t+1−1/3+t+44/3, then replace t by x2 and integrate each part as an inverse tangent. For ∫x4+1x2+1dx and ∫x4+1x2−1dx, divide by x2 and substitute u=x−x1 or u=x+x1 respectively. And for ∫x(xn+1)dx, multiply by xn−1 and substitute u=xn.
Key Takeaways:
- Divide first if the fraction is improper.
- Linear (x−a): x−aA; repeated (x−a)k: all powers up to k; quadratic: quadraticBx+C.
- Cover-up: A=Q1(a)R(a) for simple linear factors.
- ∫x2−a2dx=2a1ln∣x+ax−a∣.
- Functions of x2 only: decompose in t=x2.
- x4+1x2±1: divide by x2 and use x∓x1.
- Pitfall: using a constant numerator over an irreducible quadratic, or omitting lower powers of a repeated factor.
Brain-Teaser 1: Evaluate ∫(x−1)(x−2)x3dx.
Solution: The fraction is improper, so divide: x3=(x2−3x+2)(x+3)+7x−6. So the integrand is x+3+(x−1)(x−2)7x−6. Cover-up: at x=1, −11=−1; at x=2, 18=8. So the integral is 2x2+3x−ln∣x−1∣+8ln∣x−2∣+C. Skipping the division step would give meaningless partial fractions.
Brain-Teaser 2: Evaluate ∫(x−1)2(x+2)dx.
Solution: Write (x−1)2(x+2)1=x−1A+(x−1)2B+x+2C. Cover-up for the highest power: B=31 (at x=1), and C=91 (at x=−2). Compare the x2 coefficients after clearing denominators: 0=A+C, so A=−91. The integral is −91ln∣x−1∣−3(x−1)1+91ln∣x+2∣+C. The repeated factor needs both the first and second powers, and cover-up only finds the top one.
Brain-Teaser 3: Evaluate ∫x4+1x2+1dx.
Solution: Divide the numerator and denominator by x2: ∫x2+x211+x21dx. Since x2+x21=(x−x1)2+2 and d(x−x1)=(1+x21)dx, put u=x−x1: ∫u2+2du=21tan−12u+C=21tan−12xx2−1+C. Factoring x4+1 into real quadratics would also work but is far longer; the reciprocal substitution exploits the symmetry of the integrand.
