01 · Check the decomposition
Combine 2/(x + 1) + 3/(x − 2).
Hint
Use the common denominator (x + 1)(x − 2).
Worked solution
[2(x − 2) + 3(x + 1)]/[(x + 1)(x − 2)] = (5x − 1)/[(x + 1)(x − 2)].
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Integration using partial fractions
Integrate rational functions using partial fractions. Handle distinct and repeated linear factors, divide improper fractions, keep logarithmic scale factors and evaluate exact definite integrals.
Before you startPartial fractions, algebraic division, logarithmic integrals and the power rule.
01 / Split the fraction before integrating
A difficult quotient can become a sum of standard integrals.
Integrate the decomposed terms separately
First powers of linear factors give logarithms. Higher reciprocal powers use the power rule.
Factor the denominator, divide first if the fraction is improper, and find the partial fractions. Then integrate each term with its correct scale factor. Check the decomposition and the final derivative as two separate steps.
For the algebra itself, review partial fractions and repeated factors.
Integrate each decomposed termExplore
At x = 3: original fraction = 3.5; first term = 0.5; second term = 3; sum = 3.5.
The derivative of the combined primitive is 3.5, matching the original fraction.
Excluded points: x = −1 and x = 2.
The algebraic identity is valid wherever the original fraction is defined. A numerical check at one point illustrates it but does not prove it.
02 / Decompose distinct linear factors
Match the numerator after clearing denominators.
Find ∫(5x − 1)/[(x + 1)(x − 2)] dx.Worked example
(5x − 1)/[(x + 1)(x − 2)] = A/(x + 1) + B/(x − 2)
Use one term for each distinct linear factor.
5x − 1 = A(x − 2) + B(x + 1)
This is a polynomial identity.
x = −1 gives A = 2; x = 2 gives B = 3
These substitutions determine coefficients in the cleared identity, not values of the undefined original fraction.
2ln|x + 1| + 3ln|x − 2| + C
Integrate separately on intervals avoiding −1 and 2.
03 / Carry the linear scale into the logarithm
Integrating A/(mx + c) gives A/m times the logarithm.
Find ∫(7x + 8)/[(2x + 1)(x + 2)] dx.Worked example
3/(2x + 1) + 2/(x + 2)
Check the numerator: 3(x + 2) + 2(2x + 1) = 7x + 8.
∫3/(2x + 1) dx = (3/2)ln|2x + 1|
The derivative of 2x + 1 is 2.
(3/2)ln|2x + 1| + 2ln|x + 2| + C
The second factor has derivative 1.
02 · A scale-factor check
Integrate 5/(3x − 2).
Hint
Divide the coefficient by 3.
Worked solution
(5/3)ln|3x − 2| + C, away from x = 2/3.
03 · Repair a missing factor
A student integrates 4/(2x + 3) as 4ln|2x + 3|. Correct it.
Hint
The proposed derivative is twice the integrand.
Worked solution
2ln|2x + 3| + C.
04 / Include every power of a repeated factor
A reciprocal square does not integrate to a logarithm.
Find ∫(3x − 1)/(x − 1)² dx.Worked example
A/(x − 1) + B/(x − 1)²
Include both the first and second powers.
3x − 1 = A(x − 1) + B
Matching coefficients gives A = 3 and B = 2.
∫[3/(x − 1) + 2(x − 1)⁻²] dx
Separate the logarithmic and power terms.
3ln|x − 1| − 2/(x − 1) + C
The reciprocal-square term gains a minus sign.
Watch: logarithm and reciprocal primitives
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 · Reciprocal cube
Find ∫6/(x + 2)³ dx.
Hint
Use the exponent −3.
Worked solution
−3/(x + 2)² + C, away from x = −2.
05 · Repeated scaled factor
Find ∫4/(2x − 1)² dx.
Hint
Let u = 2x − 1; dx = du/2.
Worked solution
−2/(2x − 1) + C, away from x = 1/2.
05 / Combine repeated and distinct factors
Integrate the full decomposition term by term.
Integrate 2/(x − 1) − 3/(x − 1)² + 1/(x + 2).Worked example
∫2/(x − 1) dx = 2ln|x − 1|
The first power gives a logarithm.
∫−3(x − 1)⁻² dx = 3/(x − 1)
The two negative signs cancel.
∫1/(x + 2) dx = ln|x + 2|
This is another first-power term.
2ln|x − 1| + 3/(x − 1) + ln|x + 2| + C
Use intervals avoiding x = 1 and x = −2.
The same decomposition corresponds to (3x² − 3x − 9)/[(x − 1)²(x + 2)]. Combining it first is an independent check on the algebra.
06 · Separate integration rules
Integrate 1/(x + 1) + 2/(x + 1)².
Hint
Use a logarithm for the first term and a power for the second.
Worked solution
ln|x + 1| − 2/(x + 1) + C.
07 · Choose the full repeated-factor form
What form is needed for a proper fraction with denominator (x − 2)³(x + 1)?
Hint
Include each positive power up to the repeated multiplicity.
Worked solution
A/(x − 2) + B/(x − 2)² + C/(x − 2)³ + D/(x + 1).
06 / Divide an improper fraction first
A polynomial part must not be lost.
Find ∫(x² + 1)/(x² − 1) dx.Worked example
(x² + 1)/(x² − 1) = 1 + 2/(x² − 1)
Divide because the numerator and denominator have equal degree.
2/[(x − 1)(x + 1)] = 1/(x − 1) − 1/(x + 1)
Decompose the proper remainder.
∫[1 + 1/(x − 1) − 1/(x + 1)] dx
Keep the polynomial term 1.
x + ln|x − 1| − ln|x + 1| + C
Differentiate to check the original quotient.
08 · Identify the polynomial part
Rewrite (x² + 4x + 5)/(x + 1) by division.
Hint
Multiply x + 1 by x + 3.
Worked solution
x + 3 + 2/(x + 1).
09 · Integrate after division
Integrate (x² + 4x + 5)/(x + 1).
Hint
Use the decomposition from the previous question.
Worked solution
x²/2 + 3x + 2ln|x + 1| + C, away from x = −1.
07 / Combine logarithms only when it helps
Absolute values preserve valid negative branches.
Simplify the logarithmic part of the improper-fraction answer.Worked example
ln|x − 1| − ln|x + 1|
Each logarithm is real when its argument is nonzero.
ln|(x − 1)/(x + 1)|
Use the quotient law with absolute values.
x + ln|(x − 1)/(x + 1)| + C
The combined form retains the same excluded points ±1.
Keeping two logarithms is also correct. Do not erase absolute values merely to make the answer look shorter. On a specified interval where an argument is known positive, they may be removed with justification.
10 · A log-law correction
Is 2ln|x + 1| + 3ln|x − 2| equal to ln|2(x + 1) + 3(x − 2)|?
Hint
Logarithmic addition corresponds to multiplication, not addition of arguments.
Worked solution
No. A valid combined expression is ln|(x + 1)²(x − 2)³|, on the original domain.
08 / Check the interval before evaluating bounds
Exact logarithmic answers need no decimal approximation.
Evaluate ∫₃⁴ (5x − 1)/[(x + 1)(x − 2)] dx.Worked example
No pole lies in [3,4]
The excluded points are −1 and 2.
[2ln(x + 1) + 3ln(x − 2)]₃⁴
Both logarithm arguments are positive here.
2ln(5/4) + 3ln 2
Subtract the lower values.
ln(25/2)
Combine if desired; either exact form is acceptable.
11 · A repeated-factor definite integral
Evaluate ∫₂³ (3x − 1)/(x − 1)² dx.
Hint
Use 3ln|x − 1| − 2/(x − 1).
Worked solution
(3ln 2 − 1) − (−2) = 3ln 2 + 1.
12 · An improper fraction with valid bounds
Evaluate ∫₂³ (x² + 1)/(x² − 1) dx.
Hint
Use x + ln|(x − 1)/(x + 1)|.
Worked solution
1 + ln(3/2).
09 / Do not cross a pole
Decomposition does not change where the original integrand is defined.
For the distinct-factor example, an interval crossing x = 2 includes a pole. Subtracting finite endpoint logarithms would miss the divergence. Work on a connected interval that avoids all denominator zeros.
A factor cancellation may produce a removable hole rather than a pole, but the original expression still has its stated domain. Treat any extension explicitly; do not silently claim the original expression was defined there.
13 · Reject an invalid definite shortcut
Can the distinct-factor primitive be used directly to evaluate the ordinary integral from x = 1 to x = 3?
Hint
Locate x = 2.
Worked solution
No. The pole at x = 2 lies inside the interval and the ordinary improper integral does not converge.
10 / Verify the algebra and calculus separately
Two checks catch two different kinds of mistake.
First recombine the partial fractions to recover the original numerator and denominator. Then differentiate the proposed primitive to recover the decomposed sum. A correct integration cannot repair an incorrect decomposition.
Watch for a missed polynomial quotient, an omitted repeated-factor term, a missing linear scale, or a logarithm used where a negative power was required.
14 · Distinguish the two rules
Which of ∫1/(x − 1) dx and ∫1/(x − 1)² dx gives a logarithm?
Hint
Only the exponent −1 is the logarithmic case.
Worked solution
The first is ln|x − 1| + C; the second is −1/(x − 1) + C.
15 · Check a reciprocal-square sign
Differentiate 3/(x − 1).
Hint
Write it as 3(x − 1)⁻¹.
Worked solution
−3/(x − 1)².
11 / Divide, decompose, integrate, verify
Keep the original domain throughout.
- Divide first if the rational function is improper.
- Include every required repeated-factor term.
- Find coefficients and recombine to check them.
- Use logarithms for reciprocal first powers and powers for higher ones.
- Include linear scale factors and valid absolute values.
- Check definite intervals for poles before using bounds.
16 · A final two-term integral
Integrate 2/(3x + 1) − 5/(x − 4)².
Hint
The first term needs a factor 1/3; the second uses a power.
Worked solution
(2/3)ln|3x + 1| + 5/(x − 4) + C, separately on intervals avoiding −1/3 and 4.
Section 1 of 11 · Split the fraction before integrating