Difficulty: Intermediate | Prerequisites: Partial Fractions core method and four decomposition cases (see Part 1 notes), basic ln and arctan integration
This document walks through the actual computation of several partial fraction problems, including definite integrals and problems mixing linear with quadratic factors. It also covers polynomial long division for improper fractions. These are the kinds of problems that appear on Calc 2 exams, worked step by step so you can see where mistakes tend to happen and how to avoid them.
Each worked example follows the same pattern: factor, set up the decomposition, solve for constants, integrate. The tricky parts are handling definite integrals (watch your ln arithmetic at the bounds), quadratic factors (which produce arctan or ln terms with a linear numerator), and knowing when to use long division before you start.
Evaluate the integral of 3/(x^2 - x - 2) dx from 0 to 1.
x^2 - x - 2 = (x + 1)(x - 2)
Both factors are linear and distinct, so this is Case 1.
3 / [(x + 1)(x - 2)] = A/(x + 1) + B/(x - 2)
Multiply both sides by (x + 1)(x - 2):
3 = A(x - 2) + B(x + 1)
Substitute x = -1: 3 = A(-3) + B(0), so A = -1
Substitute x = 2: 3 = A(0) + B(3), so B = 1
The integral becomes: integral of [-1/(x + 1) + 1/(x - 2)] dx from 0 to 1
= [-ln|x + 1| + ln|x - 2|] evaluated from 0 to 1
At x = 1: -ln|2| + ln|-1| = -ln 2 + ln 1 = -ln 2
At x = 0: -ln|1| + ln|-2| = 0 + ln 2 = ln 2
Result: (-ln 2) - (ln 2) = -2 ln 2 = ln(1/4)
Evaluate the integral of 6/(x^3 + 4x) dx from 1 to 2.
x^3 + 4x = x(x^2 + 4)
Here x^2 + 4 is an irreducible quadratic (no real roots), so this is Case 3.
1/[x(x^2 + 4)] = A/x + (Bx + C)/(x^2 + 4)
Note the factor of 6 is pulled out front as a constant multiplier.
Multiply through by x(x^2 + 4):
1 = A(x^2 + 4) + x(Bx + C)
Expand: 1 = Ax^2 + 4A + Bx^2 + Cx = (A + B)x^2 + Cx + 4A
Matching coefficients:
x^2 term: A + B = 0
x^1 term: C = 0
x^0 term: 4A = 1, so A = 1/4
From A + B = 0: B = -1/4. And C = 0.
So 1/[x(x^2 + 4)] = (1/4)/x + (-x/4)/(x^2 + 4) = (1/4)/x - x/[4(x^2 + 4)]
6 times the integral of [(1/4)/x - x/(4(x^2 + 4))] dx
The first term: (6/4) ln|x| = (3/2) ln|x|
The second term: use substitution u = x^2 + 4, du = 2x dx. This gives -(6/4)(1/2) ln|x^2 + 4| = -(3/4) ln(x^2 + 4)
Evaluating from 1 to 2:
[(3/2) ln 2 - (3/4) ln 8] - [(3/2) ln 1 - (3/4) ln 5]
= (3/2) ln 2 - (3/4) ln 8 + (3/4) ln 5
= (3/2) ln 2 - (3/4)(3 ln 2) + (3/4) ln 5
= (3/2) ln 2 - (9/4) ln 2 + (3/4) ln 5
= -(3/4) ln 2 + (3/4) ln 5
= (3/4) ln(5/2)
Decompose 1/[(x - 1)(x^2 + 1)].
1/[(x - 1)(x^2 + 1)] = A/(x - 1) + (Bx + C)/(x^2 + 1)
The linear factor (x - 1) gets a constant numerator. The irreducible quadratic (x^2 + 1) gets a linear numerator (Bx + C).
Multiply both sides by (x - 1)(x^2 + 1):
1 = A(x^2 + 1) + (Bx + C)(x - 1)
Expand: 1 = Ax^2 + A + Bx^2 - Bx + Cx - C
Group: 1 = (A + B)x^2 + (-B + C)x + (A - C)
x^2: A + B = 0
x^1: -B + C = 0
x^0: A - C = 1
From x^0: A - C = 1. From x^1: B = C. From x^2: A = -B = -C.
Substituting A = -C into A - C = 1 gives -2C = 1, so C = -1/2.
Then B = -1/2 and A = 1/2.
1/[(x - 1)(x^2 + 1)] = (1/2)/(x - 1) + (-x/2 - 1/2)/(x^2 + 1)
The integral of this produces (1/2) ln|x - 1| - (1/4) ln(x^2 + 1) - (1/2) arctan(x) + C.
Factor: x^2 - 3x + 2 = (x - 1)(x - 2)
Decompose: (3x + 1)/[(x - 1)(x - 2)] = A/(x - 1) + B/(x - 2)
Multiply through: 3x + 1 = A(x - 2) + B(x - 1)
Expand: 3x + 1 = (A + B)x + (-2A - B)
From A + B = 3 and -2A - B = 1:
Adding the equations: -A = 4, so A = -4. Then B = 7.
Result: (3x + 1)/[(x - 1)(x - 2)] = -4/(x - 1) + 7/(x - 2)
Factor the denominator by grouping: x^2(x - 1) + 2(x - 1) = (x - 1)(x^2 + 2)
Decompose: (x^2 + x + 1)/[(x - 1)(x^2 + 2)] = A/(x - 1) + (Bx + C)/(x^2 + 2)
Multiply through: x^2 + x + 1 = A(x^2 + 2) + (Bx + C)(x - 1)
Expand: x^2 + x + 1 = Ax^2 + 2A + Bx^2 - Bx + Cx - C
Group: x^2 + x + 1 = (A + B)x^2 + (-B + C)x + (2A - C)
Equating coefficients: A + B = 1, -B + C = 1, 2A - C = 1
Solving this system gives the constants needed for integration.
If the degree of P(x) is greater than or equal to the degree of Q(x), the rational function P(x)/Q(x) is improper. You cannot apply partial fraction decomposition directly. First, perform polynomial long division to rewrite the fraction as:
P(x)/Q(x) = S(x) + R(x)/Q(x)
where S(x) is the quotient polynomial and R(x) is the remainder, which has degree less than Q(x). Only R(x)/Q(x) needs partial fraction decomposition.
Simplify f(x) = (x^2 - x)/(x^2 + 3x + 5).
Both numerator and denominator are degree 2, so the fraction is improper.
Divide x^2 - x + 0 by x^2 + 3x + 5:
x^2 + 3x + 5 goes into x^2 - x + 0 exactly 1 time
Multiply: 1 times (x^2 + 3x + 5) = x^2 + 3x + 5
Subtract: (x^2 - x + 0) - (x^2 + 3x + 5) = -4x - 5
Result: (x^2 - x)/(x^2 + 3x + 5) = 1 + (-4x - 5)/(x^2 + 3x + 5)
Now only the remainder fraction (-4x - 5)/(x^2 + 3x + 5) would need further decomposition if the denominator factored. Since x^2 + 3x + 5 has a negative discriminant (9 - 20 = -11), it is irreducible, and you would integrate it using completing the square and the arctan formula.
Always compare degrees before setting up partial fractions. If top degree >= bottom degree, do long division first. This check takes five seconds and saves you from a wrong answer.
When evaluating a definite integral involving ln terms, students frequently drop absolute value signs or make sign errors when substituting bounds. Write out each evaluation step explicitly.
Students sometimes forget that ln|a| - ln|b| = ln|a/b|, not ln|a - b|. Log rules matter here.
In problems with an irreducible quadratic, students try to use strategic substitution by plugging in complex roots. Stick to equating coefficients for quadratic factors.
When long division gives a remainder, students sometimes forget to integrate the quotient polynomial S(x) as well. The full integral is the integral of S(x) plus the integral of R(x)/Q(x).
Definite integral problems with partial fractions are an exam favourite. Practise the full pipeline: decompose, integrate, evaluate at bounds, simplify the ln expression.
Problems that combine long division with partial fractions test two skills at once. Expect at least one on a comprehensive exam.
The arctan integral that arises from irreducible quadratic factors is commonly tested. Know that the integral of 1/(x^2 + a^2) dx = (1/a) arctan(x/a) + C.
Exam graders look for correct decomposition setup. Partial credit often depends on whether you wrote the right form, even if the algebra goes wrong after that.
Q: Evaluate the integral of 6/(x^2 - 1) dx.
A: Factor: (x - 1)(x + 1). Decompose: 6/[(x - 1)(x + 1)] = A/(x - 1) + B/(x + 1). Solving gives A = 3, B = -3. Integral: 3 ln|x - 1| - 3 ln|x + 1| + C = 3 ln|(x - 1)/(x + 1)| + C.
Q: Set up the partial fraction decomposition for x/[(x - 1)(x - 1)], that is, x/(x - 1)^2.
A: This is a repeated linear factor. The decomposition is A/(x - 1) + B/(x - 1)^2. Multiply through: x = A(x - 1) + B. Expanding: x = Ax + (-A + B). So A = 1 and B = 1.
Q: Perform long division on (2x^3 + x)/(x^2 + 1) and identify the proper remainder fraction.
A: Divide 2x^3 + x by x^2 + 1. First term: 2x. Multiply: 2x(x^2 + 1) = 2x^3 + 2x. Subtract: (2x^3 + x) - (2x^3 + 2x) = -x. Result: 2x + (-x)/(x^2 + 1). The remainder fraction -x/(x^2 + 1) is proper and integrates to -(1/2) ln(x^2 + 1) + C.
Q: What is the integral of 1/[x(x^2 + 4)] dx?
A: Decompose as (1/4)/x - x/[4(x^2 + 4)] (from Worked Example 2). Integrate: (1/4) ln|x| - (1/8) ln(x^2 + 4) + C.
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