How to integrate any quotient of polynomials by dividing first when needed, splitting into partial fractions, and integrating each simple piece on intervals where the denominator is not zero.
A fraction like x2−2x−35x−3 looks hard to integrate. Yet it splits as
x2−2x−35x−3=x+12+x−33,
and each of those integrates to a logarithm at once. The method of partial fractions finds such splittings for every rational function, so every rational function can be integrated with polynomials, rational functions, logarithms and arctangents.
Key terms
Proper and improper rational functions
Definition A rational function is a quotient Q(x)P(x) of polynomials with Q not the zero polynomial. It is proper if degP<degQ and improper if degP≥degQ.
In plain words A proper rational function has a top of lower degree than its bottom, like a proper fraction such as 72. Partial fractions only work on proper ones.
Example The function x2+1x+4 is proper. The function x−1x3+x is improper, with degrees 3 and 1, and so is x2+1x2.
Polynomial division
Definition For polynomials P and Q with Q not zero, there are unique polynomials S and R with
P(x)=S(x)Q(x)+R(x),R=0ordegR<degQ.
Then QP=S+QR at every x with Q(x)=0, where S is a polynomial and QR is proper.
In plain words Long division splits an improper fraction into a polynomial, which is easy to integrate, plus a proper remainder fraction.
Definition Every nonconstant polynomial Q with real coefficients factors into a constant times linear factors (ax+b) with a=0 and irreducible quadratic factors (ax2+bx+c) with a=0 and b2−4ac<0. If QP is proper, it equals a sum of partial fractions, uniquely determined, made of the following terms. Each linear factor (ax+b) appearing exactly to the power r contributes
ax+bA1+(ax+b)2A2+⋯+(ax+b)rAr.
Each irreducible quadratic factor (ax2+bx+c) appearing exactly to the power s contributes
ax2+bx+cB1x+C1+⋯+(ax2+bx+c)sBsx+Cs.
The Ai, Bj and Cj are real constants.
In plain words Factor the denominator, write one simple fraction for each factor (and for each of its powers), then find the unknown numerators.
Example The denominator x2−1=(x−1)(x+1) has two distinct linear factors, and
x2−11=x−11/2−x+11/2,∫x2−1dx=21lnx+1x−1+C.
Coefficient matching
Definition To find the constants, multiply both sides of the decomposition by Q(x). This gives an identity between two polynomials that holds for all x (first for x with Q(x)=0, then for all x by continuity). Two polynomials are equal for all x exactly when the coefficients of each power of x agree. Equating coefficients, or substituting convenient values of x such as the roots of Q, gives linear equations for the constants.
In plain words Clear the denominators, then either compare the coefficients on both sides or plug in numbers that make terms vanish.
Example Write
(x+1)(x−3)5x−3=x+1A+x−3B.
Clearing denominators gives
5x−3=A(x−3)+B(x+1).
Matching the x coefficients gives A+B=5 and matching constants gives −3A+B=−3. Subtracting, 4A=8, so A=2 and B=3. As a check, x=3 gives 12=4B and x=−1 gives −8=−4A. So
∫(x+1)(x−3)5x−3dx=2ln∣x+1∣+3ln∣x−3∣+C.
Repeated linear factors
Definition A linear factor (ax+b) that appears to the power r≥2 needs all r terms (ax+b)kAk for k=1,…,r. For k≥2 they integrate to powers,
∫(ax+b)kdx=−a(k−1)(ax+b)k−11+C,a=0.
In plain words A squared factor needs two fractions, one over the factor and one over its square. Leaving out the lower power usually makes the coefficients impossible to match.
Example For x2(x−2)4 write
x2(x−2)4=xA+x2B+x−2C,
so 4=Ax(x−2)+B(x−2)+Cx2. Setting x=0 gives B=−2, setting x=2 gives C=1, and the x2 coefficients give A+C=0, so A=−1.
∫x2(x−2)4dx=−ln∣x∣+x2+ln∣x−2∣+C.
Irreducible quadratic factors
Definition A quadratic ax2+bx+c with real coefficients is irreducible over the reals when b2−4ac<0, so it has no real roots. Its partial fractions have linear numerators Bx+C. A term over the first power of the quadratic is integrated by splitting off a multiple of the derivative of the quadratic, which gives a logarithm, and completing the square in the rest, which gives an arctangent by
∫x2+k2dx=k1arctankx+C,k>0.
In plain words A quadratic that cannot be factored gets a numerator of the form Bx+C. Its integral is a logarithm part plus an arctangent part.
Example Write
x(x2+1)1=xA+x2+1Bx+C.
Clearing denominators gives 1=A(x2+1)+(Bx+C)x. Then x=0 gives A=1, the x2 coefficients give A+B=0, so B=−1, and the x coefficients give C=0.
∫x(x2+1)dx=∫(x1−x2+1x)dx=ln∣x∣−21ln(x2+1)+C.
Real-domain intervals
Definition An antiderivative of QP found this way is valid on each open interval on which Q has no zero. On different such intervals the constants of integration may differ. Suppose QP is in lowest terms, so P and Q have no common factor. A definite integral ∫abQ(x)P(x)dx can be evaluated with the fundamental theorem only if Q has no zero in [a,b]. Otherwise the integral is improper and needs separate treatment.
In plain words The logarithms carry absolute values so the answer works on both sides of each zero of the denominator. But you may never integrate across a zero of the denominator by plugging in.
Example On [2,3] the function x2−11 is continuous, so
∫23x2−1dx=21(ln42−ln31)=21ln23≈0.203.
The same formula cannot be used on [0,2], which contains the zero x=1 of the denominator.
Check yourself
Find ∫x2+2x+5dx.
Show the answer
Completing the square gives x2+2x+5=(x+1)2+4, which has no real roots. With u=x+1 and k=2 the result is
∫x2+2x+5dx=21arctan2x+1+C.
Find ∫x+1x2dx.
Show the answer
Division gives x2=(x+1)(x−1)+1, so the integrand is x−1+x+11 and
∫x+1x2dx=2x2−x+ln∣x+1∣+C.
Evaluate ∫01(x+1)(x−3)5x−3dx.
Show the answer
The denominator has no zero in [0,1]. Using the decomposition from above, the value is
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