Note: this is a BC Calculus topic. AB students can skip this section.
What is a Rational Function? A rational function has the form
P ( x ) Q ( x ) \frac{P(x)}{Q(x)} Q ( x ) P ( x ) where both
P ( x ) P(x) P ( x ) and
Q ( x ) Q(x) Q ( x ) are polynomials.
Proper vs. Improper Rational Functions A rational function is proper if
degree of P ( x ) < degree of Q ( x ) \text{degree of } P(x) < \text{degree of } Q(x) degree of P ( x ) < degree of Q ( x ) Proper example: x + 3 x 2 + 5 x + 6 \dfrac{x + 3}{x^2 + 5x + 6} x 2 + 5 x + 6 x + 3 (degree 1 over degree 2)
Improper example: x 3 + 2 x 2 − 1 \dfrac{x^3 + 2}{x^2 - 1} x 2 − 1 x 3 + 2 (degree 3 over degree 2)
You can only use partial fraction decomposition on proper rational functions.
If the fraction is improper, do long division first (Topic 6.10), then use partial fractions on the proper fraction that’s left over. For the improper example above:
x 3 + 2 x 2 − 1 = x + x + 2 x 2 − 1 \frac{x^3 + 2}{x^2 - 1} = x + \frac{x + 2}{x^2 - 1} x 2 − 1 x 3 + 2 = x + x 2 − 1 x + 2 The Big Idea Some rational functions are hard to integrate directly, but if you can break them apart into simpler fractions, the integration becomes easy.
Partial fraction decomposition rewrites one complicated fraction as a sum of simpler ones.
General Form P ( x ) Q ( x ) = A 1 F 1 ( x ) + A 2 F 2 ( x ) + A 3 F 3 ( x ) + … \frac{P(x)}{Q(x)} = \frac{A_1}{F_1(x)} + \frac{A_2}{F_2(x)} + \frac{A_3}{F_3(x)} + \ldots Q ( x ) P ( x ) = F 1 ( x ) A 1 + F 2 ( x ) A 2 + F 3 ( x ) A 3 + … where each
F i ( x ) F_i(x) F i ( x ) is a factor of
Q ( x ) Q(x) Q ( x ) .
Non-Repeating Linear Factors In BC Calculus, we focus on
non-repeating linear factors , meaning factors of the form
( x − a ) (x - a) ( x − a ) that show up only once.
The Method If
Q ( x ) Q(x) Q ( x ) factors completely into distinct linear factors,
Q ( x ) = ( x − r 1 ) ( x − r 2 ) ( x − r 3 ) ⋯ ( x − r n ) Q(x) = (x - r_1)(x - r_2)(x - r_3)\cdots(x - r_n) Q ( x ) = ( x − r 1 ) ( x − r 2 ) ( x − r 3 ) ⋯ ( x − r n ) then
P ( x ) Q ( x ) = A 1 x − r 1 + A 2 x − r 2 + A 3 x − r 3 + … + A n x − r n \frac{P(x)}{Q(x)} = \frac{A_1}{x - r_1} + \frac{A_2}{x - r_2} + \frac{A_3}{x - r_3} + \ldots + \frac{A_n}{x - r_n} Q ( x ) P ( x ) = x − r 1 A 1 + x − r 2 A 2 + x − r 3 A 3 + … + x − r n A n The goal is to find the constants
A 1 , A 2 , A 3 , … , A n A_1, A_2, A_3, \ldots, A_n A 1 , A 2 , A 3 , … , A n .
Step-by-Step Process Step 1: Factor the denominator. Factor
Q ( x ) Q(x) Q ( x ) completely into linear factors.
Step 2: Set up the partial fractions. Write the equation with unknown constants in the numerators.
Step 3: Clear the denominators. Multiply both sides by
Q ( x ) Q(x) Q ( x ) to get rid of all the fractions.
Step 4: Solve for the constants. Strategically pick values of
x x x that make certain terms vanish, then solve for each constant.
Step 5: Rewrite the integral. Put the constants back in and integrate term by term.
Example 1: Two Linear Factors Evaluate
∫ 3 ( x − 2 ) ( x + 5 ) d x \displaystyle\int \frac{3}{(x - 2)(x + 5)} \, dx ∫ ( x − 2 ) ( x + 5 ) 3 d x .
Step 1: Factor. It’s already factored:
Q ( x ) = ( x − 2 ) ( x + 5 ) Q(x) = (x - 2)(x + 5) Q ( x ) = ( x − 2 ) ( x + 5 ) Step 2: Set up the partial fractions.
3 ( x − 2 ) ( x + 5 ) = A x − 2 + B x + 5 \frac{3}{(x - 2)(x + 5)} = \frac{A}{x - 2} + \frac{B}{x + 5} ( x − 2 ) ( x + 5 ) 3 = x − 2 A + x + 5 B Step 3: Clear the denominators by multiplying both sides by
( x − 2 ) ( x + 5 ) (x - 2)(x + 5) ( x − 2 ) ( x + 5 ) :
3 = A ( x + 5 ) + B ( x − 2 ) 3 = A(x + 5) + B(x - 2) 3 = A ( x + 5 ) + B ( x − 2 ) Step 4: Solve for the constants.
Let
x = 2 x = 2 x = 2 (which kills off
B B B ):
3 = A ( 2 + 5 ) + B ( 0 ) 3 = A(2 + 5) + B(0) 3 = A ( 2 + 5 ) + B ( 0 ) 3 = 7 A 3 = 7A 3 = 7 A A = 3 7 A = \frac{3}{7} A = 7 3 Let
x = − 5 x = -5 x = − 5 (which kills off
A A A ):
3 = A ( 0 ) + B ( − 5 − 2 ) 3 = A(0) + B(-5 - 2) 3 = A ( 0 ) + B ( − 5 − 2 ) 3 = − 7 B 3 = -7B 3 = − 7 B B = − 3 7 B = -\frac{3}{7} B = − 7 3 Step 5: Rewrite and integrate.
∫ 3 ( x − 2 ) ( x + 5 ) d x = ∫ ( 3 / 7 x − 2 − 3 / 7 x + 5 ) d x \int \frac{3}{(x - 2)(x + 5)} \, dx = \int \left(\frac{3/7}{x - 2} - \frac{3/7}{x + 5}\right) dx ∫ ( x − 2 ) ( x + 5 ) 3 d x = ∫ ( x − 2 3/7 − x + 5 3/7 ) d x = 3 7 ln ∣ x − 2 ∣ − 3 7 ln ∣ x + 5 ∣ + C = 3 7 [ ln ∣ x − 2 ∣ − ln ∣ x + 5 ∣ ] + C = \frac{3}{7} \ln|x - 2| - \frac{3}{7} \ln|x + 5| + C = \frac{3}{7}\left[\ln|x - 2| - \ln|x + 5|\right] + C = 7 3 ln ∣ x − 2∣ − 7 3 ln ∣ x + 5∣ + C = 7 3 [ ln ∣ x − 2∣ − ln ∣ x + 5∣ ] + C Using log properties:
= 3 7 ln ∣ x − 2 x + 5 ∣ + C = \frac{3}{7} \ln\left|\frac{x - 2}{x + 5}\right| + C = 7 3 ln x + 5 x − 2 + C Example 2: Three Linear Factors Evaluate
∫ x 2 + 5 x 3 − 4 x 2 + 3 x d x \displaystyle\int \frac{x^2 + 5}{x^3 - 4x^2 + 3x} \, dx ∫ x 3 − 4 x 2 + 3 x x 2 + 5 d x .
Step 1: Factor the denominator.
x 3 − 4 x 2 + 3 x = x ( x 2 − 4 x + 3 ) = x ( x − 1 ) ( x − 3 ) x^3 - 4x^2 + 3x = x(x^2 - 4x + 3) = x(x - 1)(x - 3) x 3 − 4 x 2 + 3 x = x ( x 2 − 4 x + 3 ) = x ( x − 1 ) ( x − 3 ) Step 2: Set up the partial fractions.
x 2 + 5 x ( x − 1 ) ( x − 3 ) = A x + B x − 1 + C x − 3 \frac{x^2 + 5}{x(x - 1)(x - 3)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x - 3} x ( x − 1 ) ( x − 3 ) x 2 + 5 = x A + x − 1 B + x − 3 C Step 3: Clear the denominators by multiplying by
x ( x − 1 ) ( x − 3 ) x(x - 1)(x - 3) x ( x − 1 ) ( x − 3 ) :
x 2 + 5 = A ( x − 1 ) ( x − 3 ) + B x ( x − 3 ) + C x ( x − 1 ) x^2 + 5 = A(x - 1)(x - 3) + Bx(x - 3) + Cx(x - 1) x 2 + 5 = A ( x − 1 ) ( x − 3 ) + B x ( x − 3 ) + C x ( x − 1 ) Step 4: Solve for the constants.
0 + 5 = A ( − 1 ) ( − 3 ) + 0 + 0 0 + 5 = A(-1)(-3) + 0 + 0 0 + 5 = A ( − 1 ) ( − 3 ) + 0 + 0 5 = 3 A ⇒ A = 5 3 5 = 3A \quad \Rightarrow \quad A = \frac{5}{3} 5 = 3 A ⇒ A = 3 5 1 + 5 = 0 + B ( 1 ) ( − 2 ) + 0 1 + 5 = 0 + B(1)(-2) + 0 1 + 5 = 0 + B ( 1 ) ( − 2 ) + 0 6 = − 2 B ⇒ B = − 3 6 = -2B \quad \Rightarrow \quad B = -3 6 = − 2 B ⇒ B = − 3 9 + 5 = 0 + 0 + C ( 3 ) ( 2 ) 9 + 5 = 0 + 0 + C(3)(2) 9 + 5 = 0 + 0 + C ( 3 ) ( 2 ) 14 = 6 C ⇒ C = 7 3 14 = 6C \quad \Rightarrow \quad C = \frac{7}{3} 14 = 6 C ⇒ C = 3 7 Step 5: Rewrite and integrate.
∫ x 2 + 5 x ( x − 1 ) ( x − 3 ) d x = ∫ ( 5 / 3 x − 3 x − 1 + 7 / 3 x − 3 ) d x \int \frac{x^2 + 5}{x(x - 1)(x - 3)} \, dx = \int \left(\frac{5/3}{x} - \frac{3}{x - 1} + \frac{7/3}{x - 3}\right) dx ∫ x ( x − 1 ) ( x − 3 ) x 2 + 5 d x = ∫ ( x 5/3 − x − 1 3 + x − 3 7/3 ) d x = 5 3 ln ∣ x ∣ − 3 ln ∣ x − 1 ∣ + 7 3 ln ∣ x − 3 ∣ + C = \frac{5}{3}\ln|x| - 3\ln|x - 1| + \frac{7}{3}\ln|x - 3| + C = 3 5 ln ∣ x ∣ − 3 ln ∣ x − 1∣ + 3 7 ln ∣ x − 3∣ + C Example 3: Requires Factoring First Evaluate
∫ 4 x + 1 x 2 + x − 6 d x \displaystyle\int \frac{4x + 1}{x^2 + x - 6} \, dx ∫ x 2 + x − 6 4 x + 1 d x .
Step 1: Factor.
x 2 + x − 6 = ( x + 3 ) ( x − 2 ) x^2 + x - 6 = (x + 3)(x - 2) x 2 + x − 6 = ( x + 3 ) ( x − 2 ) Step 2: Set up the partial fractions.
4 x + 1 ( x + 3 ) ( x − 2 ) = A x + 3 + B x − 2 \frac{4x + 1}{(x + 3)(x - 2)} = \frac{A}{x + 3} + \frac{B}{x - 2} ( x + 3 ) ( x − 2 ) 4 x + 1 = x + 3 A + x − 2 B Step 3: Clear the denominators.
4 x + 1 = A ( x − 2 ) + B ( x + 3 ) 4x + 1 = A(x - 2) + B(x + 3) 4 x + 1 = A ( x − 2 ) + B ( x + 3 ) Step 4: Solve for the constants.
4 ( − 3 ) + 1 = A ( − 5 ) + 0 4(-3) + 1 = A(-5) + 0 4 ( − 3 ) + 1 = A ( − 5 ) + 0 − 11 = − 5 A ⇒ A = 11 5 -11 = -5A \quad \Rightarrow \quad A = \frac{11}{5} − 11 = − 5 A ⇒ A = 5 11 4 ( 2 ) + 1 = 0 + B ( 5 ) 4(2) + 1 = 0 + B(5) 4 ( 2 ) + 1 = 0 + B ( 5 ) 9 = 5 B ⇒ B = 9 5 9 = 5B \quad \Rightarrow \quad B = \frac{9}{5} 9 = 5 B ⇒ B = 5 9 Step 5: Integrate.
∫ 4 x + 1 x 2 + x − 6 d x = ∫ ( 11 / 5 x + 3 + 9 / 5 x − 2 ) d x \int \frac{4x + 1}{x^2 + x - 6} \, dx = \int \left(\frac{11/5}{x + 3} + \frac{9/5}{x - 2}\right) dx ∫ x 2 + x − 6 4 x + 1 d x = ∫ ( x + 3 11/5 + x − 2 9/5 ) d x = 11 5 ln ∣ x + 3 ∣ + 9 5 ln ∣ x − 2 ∣ + C = \frac{11}{5}\ln|x + 3| + \frac{9}{5}\ln|x - 2| + C = 5 11 ln ∣ x + 3∣ + 5 9 ln ∣ x − 2∣ + C Example 4: A Definite Integral Evaluate
∫ 3 5 2 x 2 − 1 d x \displaystyle\int_3^5 \frac{2}{x^2 - 1} \, dx ∫ 3 5 x 2 − 1 2 d x .
Step 1: Factor.
x 2 − 1 = ( x − 1 ) ( x + 1 ) x^2 - 1 = (x - 1)(x + 1) x 2 − 1 = ( x − 1 ) ( x + 1 ) Steps 2 to 4: Set up the partial fractions, clear the denominators, and solve.
2 ( x − 1 ) ( x + 1 ) = A x − 1 + B x + 1 ⇒ 2 = A ( x + 1 ) + B ( x − 1 ) \frac{2}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} \quad \Rightarrow \quad 2 = A(x + 1) + B(x - 1) ( x − 1 ) ( x + 1 ) 2 = x − 1 A + x + 1 B ⇒ 2 = A ( x + 1 ) + B ( x − 1 ) Let
x = 1 x = 1 x = 1 :
2 = 2 A 2 = 2A 2 = 2 A , so
A = 1 A = 1 A = 1 . Let
x = − 1 x = -1 x = − 1 :
2 = − 2 B 2 = -2B 2 = − 2 B , so
B = − 1 B = -1 B = − 1 .
Step 5: Integrate and evaluate.
∫ 3 5 ( 1 x − 1 − 1 x + 1 ) d x = [ ln ∣ x − 1 ∣ − ln ∣ x + 1 ∣ ] 3 5 \int_3^5 \left(\frac{1}{x - 1} - \frac{1}{x + 1}\right) dx = \Big[\ln|x - 1| - \ln|x + 1|\Big]_3^5 ∫ 3 5 ( x − 1 1 − x + 1 1 ) d x = [ ln ∣ x − 1∣ − ln ∣ x + 1∣ ] 3 5 = ( ln 4 − ln 6 ) − ( ln 2 − ln 4 ) = ln 2 3 − ln 1 2 = ln 4 3 ≈ 0.288 = (\ln 4 - \ln 6) - (\ln 2 - \ln 4) = \ln\frac{2}{3} - \ln\frac{1}{2} = \ln\frac{4}{3} \approx 0.288 = ( ln 4 − ln 6 ) − ( ln 2 − ln 4 ) = ln 3 2 − ln 2 1 = ln 3 4 ≈ 0.288 Check the interval first. The integrand has vertical asymptotes at
x = 1 x = 1 x = 1 and
x = − 1 x = -1 x = − 1 . Neither is in
[ 3 , 5 ] [3, 5] [ 3 , 5 ] , so this is an ordinary definite integral. If one of them were inside the interval, it would be an improper integral (Topic 6.13).
Why No U-Substitution? When integrating terms like
∫ 1 x − 3 d x \displaystyle\int \frac{1}{x - 3} \, dx ∫ x − 3 1 d x , notice that:
The derivative of ( x − 3 ) (x - 3) ( x − 3 ) 1 1 1 There’s no coefficient and no chain rule involved, so no substitution is needed The antiderivative is simply ln ∣ x − 3 ∣ + C \ln|x - 3| + C ln ∣ x − 3∣ + C This applies to all the linear factors in partial fractions.
Logarithm Properties Review When simplifying answers, remember these:
ln ( a ) + ln ( b ) = ln ( a b ) \ln(a) + \ln(b) = \ln(ab) ln ( a ) + ln ( b ) = ln ( ab ) ln ( a ) − ln ( b ) = ln ( a b ) \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) ln ( a ) − ln ( b ) = ln ( b a ) k ln ( a ) = ln ( a k ) k\ln(a) = \ln(a^k) k ln ( a ) = ln ( a k ) Example Simplification 2 ln ∣ x ∣ − 3 ln ∣ x + 1 ∣ 2\ln|x| - 3\ln|x + 1| 2 ln ∣ x ∣ − 3 ln ∣ x + 1∣ can be rewritten as
ln ∣ x 2 ∣ − ln ∣ x + 1 ∣ 3 = ln ∣ x 2 ( x + 1 ) 3 ∣ \ln|x^2| - \ln|x + 1|^3 = \ln\left|\frac{x^2}{(x + 1)^3}\right| ln ∣ x 2 ∣ − ln ∣ x + 1 ∣ 3 = ln ( x + 1 ) 3 x 2 Some AP exam problems require this kind of simplification.
Common Mistakes to Avoid Mistake 1: Forgetting Absolute Values Always use absolute values inside logarithms:
∫ 1 x − 5 d x = ln ∣ x − 5 ∣ + C \int \frac{1}{x - 5} \, dx = \ln|x - 5| + C ∫ x − 5 1 d x = ln ∣ x − 5∣ + C not
ln ( x − 5 ) + C \ln(x - 5) + C ln ( x − 5 ) + C .
Mistake 2: Multiplying Fractions Instead of Adding Partial fractions are added , not multiplied:
1 ( x − 1 ) ( x + 2 ) = A x − 1 + B x + 2 \frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} ( x − 1 ) ( x + 2 ) 1 = x − 1 A + x + 2 B Mistake 3: Not Fully Factoring Always factor completely before setting up the partial fractions.
Strategy Summary Check if it’s proper: the numerator’s degree has to be less than the denominator’sFactor completely: break the denominator into all its linear factorsSet up the fractions: one fraction for each factorClear the denominators: multiply through by the full denominatorSubstitute strategically: pick x x x Integrate term by term: each term becomes a natural logSimplify if needed using log propertiesKey Takeaways Partial fractions break complicated rational functions into simpler pieces for integrationProper rational functions: the numerator’s degree has to be less than the denominator’sEach linear factor becomes its own fraction, A x − r \frac{A}{x - r} x − r A Clear the denominators by multiplying both sides by the factored form of Q ( x ) Q(x) Q ( x ) Substitute strategically. Pick x x x The result: each term integrates to ln ∣ x − r ∣ \ln|x - r| ln ∣ x − r ∣ Simplify with log properties: ln ( a ) ± ln ( b ) = ln ( a b ) \ln(a) \pm \ln(b) = \ln(ab) ln ( a ) ± ln ( b ) = ln ( ab ) ln ( a b ) \ln\left(\frac{a}{b}\right) ln ( b a ) BC scope: only non-repeating linear factors
