Pearson Edexcel International Advanced Level Mathematics: Integration by substitution, by parts and using partial fractions (YMA01)

This study guide covers Pearson Edexcel International Advanced Level Mathematics (YMA01), aligned to the International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019). It teaches Pure Mathematics 4 (P4) topics 6.2 and 6.3. The specification states that P3 and P4 are compulsory for IAL Mathematics and Pure Mathematics only; they are not part of IAS.

Use it with the revision notes and the practice questions. Standard integrals and integration by recognition are in the P3 integration guide, and finding partial fractions is in the partial fractions guide. The course hub is at Edexcel IAL Mathematics, the printable checklist tracks your progress, and the free diagnostics show where to start.

What this unit covers

Topic	What you must be able to do
6.2	Use a substitution to integrate simple cases such as x√(x − 2); when the integral is harder, the substitution is given
6.2	Integrate by parts, and understand it as the reverse of the product rule (substitution is the reverse of the chain rule)
6.2	Integrate ln x; apply parts more than once, as in x²eˣ and eˣ sin x
6.3	Integrate expressions that come from partial fractions, such as 2/(3x + 5) and 3/(x − 1)²
6.3	Integrate other rational expressions such as x/(x² + 5) and 2/(2x − 1)⁴

Volumes of revolution, separable differential equations and areas from parametric equations are other P4 topics, covered elsewhere. This page stays with the three methods above.

6.2 Integration by substitution

Substitution undoes the chain rule. If you can see an inner function whose derivative is (up to a constant) also present, replace the inner function with a single letter, integrate, then put the original variable back.

Method

1. Choose u (often the awkward bracket or root).
2. Differentiate u, then express dx using du.
3. Rewrite every x in the integrand using u. If x remains, solve the substitution for x.
4. Integrate with respect to u.
5. For an indefinite integral, substitute back. For a definite integral, either change the limits to u-values or substitute back and use the x-limits. Do not mix the two.

Worked example 1: indefinite integral

Find ∫ x√(x + 4) dx.

Let u = x + 4. Then x = u − 4 and dx = du.

∫ (u − 4) u^(1/2) du = ∫ (u^(3/2) − 4u^(1/2)) du
                     = (2/5)u^(5/2) − (8/3)u^(3/2) + c

Substituting back gives (2/5)(x + 4)^(5/2) − (8/3)(x + 4)^(3/2) + c. Differentiating confirms it.

Worked example 2: changing the limits

Evaluate ∫ from 0 to 2 of x/√(2x + 1) dx.

Let u = 2x + 1, so du = 2 dx, dx = ½ du and x = (u − 1)/2. When x = 0, u = 1. When x = 2, u = 5.

∫ from 1 to 5 of [(u − 1)/2] × u^(−1/2) × ½ du
 = ¼ ∫ (u^(1/2) − u^(−1/2)) du
 = ¼ [ (2/3)u^(3/2) − 2u^(1/2) ] from 1 to 5

At u = 5: (2/3)(5√5) − 2√5 = 4√5/3. At u = 1: 2/3 − 2 = −4/3. The difference is 4√5/3 + 4/3, so the value is ¼ × (4√5 + 4)/3 = (√5 + 1)/3.

Worked example 3: the substitution is given

Use u = eˣ to evaluate ∫ from 0 to ln 3 of 1/(eˣ + 1) dx.

Then du = eˣ dx, so dx = du/u. The limits become u = 1 and u = 3.

∫ from 1 to 3 of 1/(u(u + 1)) du = ∫ (1/u − 1/(u + 1)) du
                                 = [ln u − ln(u + 1)] from 1 to 3

This gives (ln 3 − ln 4) − (0 − ln 2) = ln 3 − ln 4 + ln 2, which is ln(3/2). The fraction 1/(u(u + 1)) has two linear factors, so it is split into partial fractions before integrating.

6.2 Integration by parts

The product rule says d(uv)/dx = u dv/dx + v du/dx. Integrating both sides and rearranging gives the formula:

∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx

Pick u as the factor that becomes simpler after differentiating, such as a power of x or ln x. Pick dv/dx as the factor you can integrate directly, such as eˣ, sin x or cos x. Parts has helped only if the new integral is easier than the one you began with.

Worked example 4: one application

Find ∫ x cos 3x dx.

Take u = x, dv/dx = cos 3x. Then du/dx = 1 and v = (1/3) sin 3x.

∫ x cos 3x dx = (x/3) sin 3x − ∫ (1/3) sin 3x dx
              = (x/3) sin 3x + (1/9) cos 3x + c

The answer is (x/3) sin 3x + (1/9) cos 3x + c.

Worked example 5: the integral of ln x

There is no obvious product, so write ln x as 1 × ln x. Take u = ln x and dv/dx = 1, so du/dx = 1/x and v = x.

∫ ln x dx = x ln x − ∫ x × (1/x) dx = x ln x − x + c

For a definite integral, ∫ from 1 to 4 of ln x dx = [x ln x − x] from 1 to 4 = (4 ln 4 − 4) − (0 − 1) = 4 ln 4 − 3, which is 8 ln 2 − 3.

Worked example 6: parts twice

Find ∫ x² e^(2x) dx.

First pass: u = x², dv/dx = e^(2x), so du/dx = 2x and v = ½ e^(2x).

∫ x² e^(2x) dx = ½x² e^(2x) − ∫ (x)(e^(2x)) dx

For the remaining integral, apply parts again with u = x and dv/dx = e^(2x). That gives ½x e^(2x) − ¼ e^(2x).

∫ x² e^(2x) dx = ½x² e^(2x) − ½x e^(2x) + ¼ e^(2x) + c

Factorising, the result is e^(2x)(½x² − ½x + ¼) + c. Each pass lowers the power of x by one, so x³ would need three passes.

Worked example 7: the integral returns to itself

Find ∫ e^(2x) cos x dx.

Let I stand for the integral. Take u = cos x, dv/dx = e^(2x).

I = ½ e^(2x) cos x + ½ ∫ e^(2x) sin x dx

Apply parts again to the new integral, with u = sin x and dv/dx = e^(2x):

∫ e^(2x) sin x dx = ½ e^(2x) sin x − ½ ∫ e^(2x) cos x dx = ½ e^(2x) sin x − ½ I

Substitute this back:

I = ½ e^(2x) cos x + ¼ e^(2x) sin x − ¼ I
(5/4) I = ¼ e^(2x)(2 cos x + sin x)

So I = e^(2x)(2 cos x + sin x)/5 + c. Use the same choice of u (trigonometric both times, or exponential both times) in the two passes. If you swap halfway, the integral cancels itself and you get 0 = 0.

6.3 Integrating rational expressions

Standard forms

Integrand	Result	Why
k/(ax + b)	(k/a) ln|ax + b| + c	derivative of the logarithm brings out a
k/(ax + b)ⁿ, n ≥ 2	k(ax + b)^(1−n) / (a(1 − n)) + c	treat as a power; divide by a
kx/(x² + a)	(k/2) ln(x² + a) + c	numerator is a multiple of the derivative of the denominator

For 2/(3x + 5) the result is (2/3) ln|3x + 5| + c, and for 3/(x − 1)² it is −3/(x − 1) + c. These are the first two forms. The third form, f′(x)/f(x) integrating to a logarithm, is recognition; see the P3 guide. For example, ∫ 3x/(x² + 7) dx = (3/2) ln(x² + 7) + c, because x² + 7 is always positive and needs no modulus.

Worked example 8: negative coefficients and powers

Find ∫ 8/(1 − 4x)³ dx. Raise the power by one, giving (1 − 4x)^(−2), then divide by the new power (−2) and by the coefficient of x (−4). Together these divide by 8, which cancels the 8 on top, so the result is (1 − 4x)^(−2) + c. Differentiating, −2 × (1 − 4x)^(−3) × (−4) = 8/(1 − 4x)³ confirms it.

Using partial fractions

When the denominator factorises, split the fraction first, then use the standard forms.

Worked example 9: distinct linear factors

Evaluate ∫ from 1 to 3 of (13x − 2)/((3x − 1)(x + 2)) dx.

Write (13x − 2)/((3x − 1)(x + 2)) = A/(3x − 1) + B/(x + 2), so 13x − 2 = A(x + 2) + B(3x − 1). Put x = −2: −28 = −7B, so B = 4. Put x = ⅓: 7/3 = (7/3)A, so A = 1.

∫ [1/(3x − 1) + 4/(x + 2)] dx = ⅓ ln|3x − 1| + 4 ln|x + 2|
At x = 3: ⅓ ln 8 + 4 ln 5 = ln 2 + 4 ln 5
At x = 1: ⅓ ln 2 + 4 ln 3
Difference: ⅔ ln 2 + 4 ln 5 − 4 ln 3

The value is ⅔ ln 2 + 4 ln(5/3). Note that 1/(3x − 1) integrates to ⅓ ln|3x − 1|, because the coefficient of x divides the result.

Worked example 10: a repeated factor

Find ∫ from 2 to 3 of (4x² − 3x + 5)/((x − 1)²(x + 2)) dx.

The form is A/(x − 1) + B/(x − 1)² + C/(x + 2), and the numerator satisfies

4x² − 3x + 5 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)².

Put x = 1: 6 = 3B, so B = 2. Put x = −2: 27 = 9C, so C = 3. Compare x²: 4 = A + C, so A = 1.

∫ [1/(x − 1) + 2/(x − 1)² + 3/(x + 2)] dx
   = ln(x − 1) − 2/(x − 1) + 3 ln(x + 2)
At x = 3: ln 2 − 1 + 3 ln 5.  At x = 2: 0 − 2 + 3 ln 4.
Difference: ln 2 + 1 + 3 ln 5 − 6 ln 2 = 1 + 3 ln 5 − 5 ln 2

The value is 1 + 3 ln 5 − 5 ln 2. If the numerator has the same degree as the denominator, or a higher one, divide first (see the revision notes) and integrate the polynomial part separately.

Choosing a method

Integrand looks like	Try
A bracket or root raised to a power, times x	Substitution on the bracket
Polynomial times eˣ, sin x or cos x	Parts, with the polynomial as u
ln x, alone or times a power of x	Parts, with ln x as u
Product of eˣ and sin x or cos x	Parts twice, then rearrange
A fraction with a factorised denominator	Partial fractions
A numerator that is the derivative of the denominator	Recognition: ln of the denominator

Common errors

• Forgetting to change dx. After u = 2x + 1 you need dx = ½ du, not dx = du.
• Leaving an x inside a u-integral. Solve the substitution for x and replace it.
• Using x-limits on a u-integral. Either convert the limits or substitute back first.
• Dropping the minus sign in parts: the second term is − ∫ v du/dx dx. In x² eˣ-type questions the sign alternates on each pass.
• Dividing by the wrong coefficient: ∫ 1/(2x − 1)⁴ dx has factors of 1/(−3) and 1/2 together, giving −1/(6(2x − 1)³).
• Giving a repeated factor two logarithms. The term B/(x − 1)² integrates to −B/(x − 1), not a logarithm.
• Missing + c on indefinite integrals, or leaving it on a definite one.

Next steps

Condense this into the revision notes, then test yourself with the practice questions. Tick off topics on the checklist and take a free 10-minute diagnostic when you want to see what is left.

Official syllabus

International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification Issue 3, April 2019, Pearson Education Limited. This page covers unit P4 topics 6.2 and 6.3.