Chapter 03 — Integration | Part 4: Partial Fractions and Integration by Parts

MATHEMATICS HIP-HOP ACADEMY • A-LEVEL PURE MATHEMATICS

CHAPTER 03 — INTEGRATION

PART 4

Partial Fractions and Integration by Parts

Two important integration techniques for handling products of functions and rational algebraic expressions. Learn how to recognise the correct method, show each stage of working and check your final answers.

Learning objectives

By the end of this part, you should be able to:

• derive and apply the integration by parts formula from the product rule;
• choose suitable expressions for u and dv;
• integrate products involving algebraic, exponential, trigonometric and logarithmic functions;
• apply integration by parts more than once when necessary;
• evaluate definite integrals using integration by parts;
• recognise rational functions suitable for partial fraction decomposition;
• resolve fractions with distinct and repeated linear factors;
• integrate the resulting partial fractions;
• use polynomial division when a rational fraction is improper.

SECTION A — INTEGRATION BY PARTS

Integration by parts is a technique used when an integrand is the product of two functions and direct integration or substitution is not convenient. It is especially useful when differentiating one factor makes it simpler.

1. Deriving the formula from the product rule

Recall the product rule for differentiation. If u and v are functions of x, then:

d(uv)/dx = u(dv/dx) + v(du/dx)

Rearranging gives:

u(dv/dx) = d(uv)/dx − v(du/dx)

Integrating both sides with respect to x:

∫u(dv/dx) dx = uv − ∫v(du/dx) dx

In differential notation, the formula is written:

∫u dv = uv − ∫v du

Important: Integration by parts is the reverse of the product rule. It changes one product integral into a product term minus another integral.

2. Choosing u and dv

The success of integration by parts depends on choosing u and dv carefully. A useful approach is:

• Choose u as the factor that becomes simpler when differentiated.
• Choose dv as the remaining factor that can be integrated easily.
• Differentiate u to obtain du.
• Integrate dv to obtain v.
• Substitute into ∫u dv = uv − ∫v du.

A useful guide for selecting u is the LIATE order:

Letter	Type of function	Example
L	Logarithmic	ln x
I	Inverse trigonometric	tan⁻¹x
A	Algebraic	x, x², 3x+1
T	Trigonometric	sin x, cos x
E	Exponential	eˣ

LIATE is a guide rather than an inflexible rule. The best choice is the one that makes the remaining integral simpler.

3. Worked examples: one application

Worked Example 1

Evaluate:

∫xeˣ dx

Solution

Choose:

u = x,    dv = eˣ dx

Then:

du = dx,    v = eˣ

Using the formula:

∫u dv = uv − ∫v du

∫xeˣ dx = xeˣ − ∫eˣ dx

= xeˣ − eˣ + C

Answer:

∫xeˣ dx = eˣ(x − 1) + C

Worked Example 2

Evaluate:

∫x cos x dx

Solution

Choose:

u = x,    dv = cos x dx

Therefore:

du = dx,    v = sin x

Applying the formula:

∫x cos x dx = x sin x − ∫sin x dx

= x sin x + cos x + C

Answer: x sin x + cos x + C

Worked Example 3

Evaluate:

∫ln x dx,   x > 0

Solution

Write the integral as:

∫ln x × 1 dx

Choose:

u = ln x,    dv = dx

Then:

du = 1/x dx,    v = x

Using integration by parts:

∫ln x dx = x ln x − ∫x(1/x) dx

= x ln x − ∫1 dx

= x ln x − x + C

Answer: x ln x − x + C

4. Repeated integration by parts

Some integrals require the integration by parts formula to be applied more than once. This commonly happens with a polynomial multiplied by an exponential or trigonometric function.

The polynomial becomes simpler each time it is differentiated. Eventually, its derivative becomes zero, allowing the integral to be completed.

Worked Example 4

Evaluate:

∫x²eˣ dx

First application

Choose:

u = x²,    dv = eˣ dx

Then:

du = 2x dx,    v = eˣ

Therefore:

∫x²eˣ dx = x²eˣ − 2∫xeˣ dx

Second application

From Worked Example 1:

∫xeˣ dx = eˣ(x − 1)

Substitute:

∫x²eˣ dx = x²eˣ − 2eˣ(x − 1) + C

Factor out eˣ:

= eˣ[x² − 2x + 2] + C

Answer: eˣ(x² − 2x + 2) + C

Worked Example 5

Evaluate:

∫x² sin x dx

Solution

Choose:

u = x²,    dv = sin x dx

Then:

du = 2x dx,    v = −cos x

Applying integration by parts:

I = −x² cos x + 2∫x cos x dx

From Worked Example 2:

∫x cos x dx = x sin x + cos x

Therefore:

I = −x² cos x + 2(x sin x + cos x) + C

Answer: −x² cos x + 2x sin x + 2cos x + C

5. Definite integration by parts

For definite integrals, the boundary values are applied to the product uv. The formula becomes:

∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du

There is no constant of integration in the final value of a definite integral.

Worked Example 6

Evaluate:

∫₀¹ xeˣ dx

Solution

From Worked Example 1:

∫xeˣ dx = eˣ(x − 1)

Apply the limits:

= [eˣ(x − 1)]₀¹

= e¹(1 − 1) − e⁰(0 − 1)

= 0 − (−1)

Answer: 1

Common examination errors

• Forgetting the negative sign when integrating sin x.
• Choosing u and dv without checking whether the remaining integral becomes simpler.
• Forgetting to integrate dv when finding v.
• Applying the formula but omitting the second integral.
• For definite integrals, failing to substitute both limits correctly.

SECTION B — PARTIAL FRACTIONS AND INTEGRATION

Partial fractions provide a way of expressing a rational function as a sum of simpler fractions. These simpler fractions are often much easier to integrate.

6. Understanding partial fractions

A rational function is a quotient of two polynomials. For example:

(5x + 1) / [(x + 1)(x + 2)]

If the denominator can be factorised, the fraction may be expressed as a sum of fractions with simpler denominators.

For example:

(5x + 1) / [(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)

Once A and B have been found, the original fraction can be integrated term by term.

First check the degrees. Before decomposing a rational function, check whether the degree of the numerator is less than the degree of the denominator. If the numerator degree is equal to or greater than the denominator degree, perform polynomial division first.

7. Partial fractions with distinct linear factors

When the denominator contains different linear factors, each factor is represented by a separate fraction.

For example:

1/[(x + a)(x + b)] = A/(x + a) + B/(x + b)

The constants are found by multiplying through by the original denominator and comparing coefficients or substituting suitable values of x.

Worked Example 7

Evaluate:

∫ (5x + 1)/[(x + 1)(x + 2)] dx

Solution

Resolve into partial fractions:

(5x + 1)/[(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)

Multiply through by (x + 1)(x + 2):

5x + 1 = A(x + 2) + B(x + 1)

Expand:

5x + 1 = Ax + 2A + Bx + B

Collect coefficients:

A + B = 5

2A + B = 1

Subtract the first equation from the second:

A = −4

Therefore B = 9.

So:

(5x + 1)/[(x + 1)(x + 2)] = −4/(x + 1) + 9/(x + 2)

Integrate:

= −4ln|x + 1| + 9ln|x + 2| + C

Answer: −4ln|x + 1| + 9ln|x + 2| + C

Worked Example 8

Evaluate:

∫ (3x + 7)/(x² + 3x + 2) dx

Solution

First factorise the denominator:

x² + 3x + 2 = (x + 1)(x + 2)

Write:

(3x + 7)/[(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)

Multiply through:

3x + 7 = A(x + 2) + B(x + 1)

Comparing coefficients:

A + B = 3

2A + B = 7

Subtracting gives:

A = 4,    B = −1

Hence:

3x + 7 over (x + 1)(x + 2) = 4/(x + 1) − 1/(x + 2)

Integrating:

= 4ln|x + 1| − ln|x + 2| + C

Answer: 4ln|x + 1| − ln|x + 2| + C

8. Partial fractions with repeated linear factors

When a linear factor is repeated, every power of that factor must be included in the decomposition.

For example, if the denominator contains (x − a)², write:

A/(x − a) + B/(x − a)²

For a factor repeated three times, the decomposition includes all three powers:

A/(x − a) + B/(x − a)² + D/(x − a)³

Do not omit any of the powers of a repeated factor. Leaving out a term generally makes it impossible to obtain the correct decomposition.

Worked Example 9

Evaluate:

∫ (x + 3)/(x + 1)² dx

Solution

Write the partial fraction form:

(x + 3)/(x + 1)² = A/(x + 1) + B/(x + 1)²

Multiply through by (x + 1)²:

x + 3 = A(x + 1) + B

Expand:

x + 3 = Ax + A + B

Compare coefficients of x:

A = 1

Compare constant terms:

A + B = 3

1 + B = 3, so B = 2.

Therefore:

(x + 3)/(x + 1)² = 1/(x + 1) + 2/(x + 1)²

Integrate:

∫1/(x + 1) dx + 2∫(x + 1)⁻² dx

= ln|x + 1| − 2/(x + 1) + C

Answer: ln|x + 1| − 2/(x + 1) + C

9. Irreducible quadratic factors

An irreducible quadratic is a quadratic expression that cannot be factorised into real linear factors.

When the denominator contains an irreducible quadratic, the numerator over that quadratic is generally linear. Sometimes the numerator can be separated into a multiple of the derivative of the denominator plus a constant.

Worked Example 10

Evaluate:

∫ (2x + 3)/(x² + 4) dx

Solution

Separate the numerator:

∫(2x + 3)/(x² + 4) dx

= ∫2x/(x² + 4) dx + 3∫1/(x² + 4) dx

First integral

Let u = x² + 4.

Then du = 2x dx.

Therefore:

∫2x/(x² + 4) dx = ln(x² + 4)

Second integral

Use the standard result:

∫dx/(x² + a²) = (1/a)tan⁻¹(x/a) + C

Here a = 2.

Therefore:

3∫dx/(x² + 4) = (3/2)tan⁻¹(x/2)

Combining both parts:

Answer: ln(x² + 4) + (3/2)tan⁻¹(x/2) + C

10. Improper rational fractions: division first

A rational fraction is improper when the degree of its numerator is equal to or greater than the degree of its denominator.

In this situation, perform polynomial division before attempting partial fraction decomposition.

Worked Example 11

Evaluate:

∫ (x² + 1)/(x − 1) dx

Solution

The numerator has degree 2 while the denominator has degree 1.

Divide x² + 1 by x − 1:

x² + 1 = (x − 1)(x + 1) + 2

Therefore:

(x² + 1)/(x − 1) = x + 1 + 2/(x − 1)

Now integrate each term:

∫[x + 1 + 2/(x − 1)] dx

= x²/2 + x + 2ln|x − 1| + C

Answer: x²/2 + x + 2ln|x − 1| + C

11. Choosing the correct integration method

Form of integrand	Suitable method
Product of two functions, such as xeˣ	Integration by parts
Polynomial multiplied by sine or cosine	Repeated integration by parts when needed
Logarithmic function such as ln x	Integration by parts
Proper rational fraction with factorable denominator	Partial fractions
Repeated linear factors	Partial fractions with every repeated power
Numerator degree at least denominator degree	Polynomial division first

Checking your answer Differentiate the final expression. For an indefinite integral, the derivative of your answer should reproduce the original integrand. This is one of the most reliable ways to check an integration result.

SECTION C — PRACTICE EXERCISES

Attempt the questions independently before opening the answer key. Show your choice of method, intermediate steps and final result.

Exercise A — Integration by parts

1. Evaluate ∫xe^(2x) dx.

2. Evaluate ∫x sin x dx.

3. Evaluate ∫x ln x dx, for x > 0.

4. Evaluate ∫x² cos x dx.

5. Evaluate ∫₀ᵖⁱ x cos x dx.

Exercise B — Partial fractions and integration

6. Evaluate ∫(2x + 5)/[(x + 1)(x + 3)] dx.

7. Evaluate ∫(4x + 1)/(x² − x − 2) dx.

8. Evaluate ∫(x + 2)/(x − 1)² dx.

9. Evaluate ∫(3x + 2)/(x² + 1) dx.

10. Evaluate ∫(x² + 4)/(x + 1) dx.

SECTION D — ANSWERS TO PRACTICE EXERCISES

Use these answers to check your work after attempting all the questions. For examination preparation, reproduce the full working independently.

Reveal answers 1–5: Integration by parts

1. ∫xe^(2x) dx

Answer: e^(2x)(x/2 − 1/4) + C

2. ∫x sin x dx

Answer: −x cos x + sin x + C

3. ∫x ln x dx

Answer: (x²/2)ln x − x²/4 + C

4. ∫x² cos x dx

Answer: x² sin x + 2x cos x − 2sin x + C

5. ∫₀ᵖⁱ x cos x dx

Answer: −2

Reveal answers 6–10: Partial fractions

6. ∫(2x + 5)/[(x + 1)(x + 3)] dx

Answer: (3/2)ln|x + 1| + (1/2)ln|x + 3| + C

7. ∫(4x + 1)/(x² − x − 2) dx

Answer: 3ln|x − 2| + ln|x + 1| + C

8. ∫(x + 2)/(x − 1)² dx

Answer: ln|x − 1| − 3/(x − 1) + C

9. ∫(3x + 2)/(x² + 1) dx

Answer: (3/2)ln(x² + 1) + 2tan⁻¹x + C

10. ∫(x² + 4)/(x + 1) dx

Answer: x²/2 − x + 5ln|x + 1| + C

FINAL EXAMINATION CHECKLIST

• State u, dv, du and v clearly when using integration by parts.
• Choose u so that differentiation makes it simpler where possible.
• Remember the negative sign when integrating sine.
• Apply integration by parts again when the remaining integral requires it.
• For definite integrals, substitute the upper limit and subtract the lower-limit value.
• Factorise the denominator before resolving a suitable rational fraction.
• Include every required term when a linear factor is repeated.
• Perform polynomial division before decomposing an improper fraction.
• Use absolute values in logarithmic expressions where the domain requires them.
• Differentiate your final indefinite integral to check the result.

Part 4 summary

Integration by parts is useful for products of functions, while partial fractions simplify suitable rational functions into forms that can be integrated term by term.

The central skill is recognising the structure of an integrand and selecting the method that makes the calculation manageable.

Practise both methods carefully, show every important algebraic step and verify your answers by differentiation.

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