OCR A Level Mathematics A: Techniques of integration, parts and partial fractions (H240) – Revision Notes

These notes condense the integration techniques in OCR A Level Mathematics A (H240), specification Version 3.1 (August 2026): content statements 1.08g to 1.08j and the numerical integration statement 1.09f. All are Stage 2 content, so they are A Level only. For full explanations and longer worked examples, read the study guide. Test yourself with the practice questions, and tick skills off on the printable checklist. The wider course is on the OCR A Level Mathematics hub.

Key results

Reference	Result or method
1.08g	Area = lim (δx→0) Σ f(x)δx = ∫ f(x) dx
1.09f	Trapezium rule: (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)], with h = (b − a)/n
1.08h	∫ f’(x)[f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/(n + 1) + c, for n ≠ −1
1.08h	∫ k f’(x)/f(x) dx = k ln
1.08h	Substitution: u = g(x); replace x, dx and the limits
1.08i	∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx
1.08j	Split into partial fractions, then ∫ 1/(px + q) dx = (1/p) ln

1.08g and 1.09f: sums, rectangles and trapezia

• Each strip has area f(x)δx. Add them, let δx → 0, and the limit is the definite integral.
• A limit of a sum written as lim Σ (x = 1 to 4) x² δx means ∫₁⁴ x² dx.
• Rectangles: for an increasing function, left-edge heights give a lower bound and right-edge heights an upper bound. For a decreasing function, swap them.
• Trapezium rule: ordinates are the y-values at equally spaced x-values. Strips = n, ordinates = n + 1.
• Concave curve (bends downwards): trapezium rule underestimates. Convex curve (bends upwards): it overestimates.

Reminder (bounds). Bound ∫₀⁴ √x dx with 4 rectangles of width 1. The function is increasing, with ordinates 0, 1, √2, √3, 2.

Lower (left edges):  1 × (0 + 1 + √2 + √3) ≈ 4.146
Upper (right edges): 1 × (1 + √2 + √3 + 2) ≈ 6.146

So 4.146 < ∫₀⁴ √x dx < 6.146 (the true value is 16/3 ≈ 5.333).

Method in steps (trapezium rule)

1. h = (b − a)/n
2. List x and y values in a table
3. Add first + last, then add twice the middle values
4. Multiply by h/2
5. State under- or over-estimate, with the reason

Reminder. A table gives y = 2.0, 2.6, 3.1, 3.5 at x = 0, 1, 2, 3, so h = 1. The estimate is ½[2.0 + 3.5 + 2(2.6 + 3.1)] = 8.45. The increases in y are shrinking (0.6, 0.5, 0.4), so the curve looks concave and the true area is probably larger.

1.08h: reverse chain rule and substitution

Method in steps (recognition)

1. Is the top a multiple of the derivative of the bracket or denominator?
2. If yes, write it as k × f'(x)
3. Raise the power by one and divide by the new power (or take ln|f|)
4. Multiply by k, add c, then differentiate to check

Reminder. ∫ 5x²/(x³ + 2) dx. The derivative of x³ + 2 is 3x², so the top is (5/3) × 3x². The answer is (5/3) ln|x³ + 2| + c.

Method in steps (substitution)

1. Pick u (inner bracket, square root, or the given substitution)
2. Find du/dx, so that dx = du / (du/dx)
3. Replace every x, including any factor of x left over, using x = (expression in u)
4. Change the limits to u-values
5. Integrate in u; substitute back only if the integral is indefinite

Reminder. Evaluate ∫₀¹ x(2 − x)⁵ dx with u = 2 − x. Then x = 2 − u, dx = −du, and u runs from 2 to 1.

∫₂¹ −(2 − u)u⁵ du = ∫₁² (2u⁵ − u⁶) du = [u⁶/3 − u⁷/7]₁² = 64/21 − 4/21 = 20/7

For √(a² − x²) try x = a sin θ, and use cos²θ = ½(1 + cos 2θ).

1.08i: integration by parts

Method in steps

1. Choose u: ln x, or a power of x (it gets simpler when differentiated)
2. Choose dv/dx: eˣ, sin, cos, or a power (you can integrate it)
3. Find du/dx and v
4. Write uv − ∫ v (du/dx) dx and integrate what remains
5. For x² times eˣ, sin or cos, repeat once
6. For a definite integral, apply the limits to uv as well

• ∫ ln x dx: take u = ln x and dv/dx = 1. The answer is x ln x − x + c.
• ∫ x^n ln x dx: always take u = ln x. For example, ∫ x ln x dx = (x²/2) ln x − ∫ x/2 dx = (x²/2) ln x − x²/4 + c.
• The reduction formulae method is not required.

Reminder. ∫₀^π x sin x dx. With u = x and v = −cos x: [−x cos x]₀^π + ∫₀^π cos x dx = π + [sin x]₀^π = π.

1.08j: partial fractions

• Denominators are no more complicated than (ax + b)(cx + d)² or (ax + b)(cx + d)(ex + f), with a constant or linear numerator.
• Forms: A/(ax + b) + B/(cx + d), or A/(ax + b) + B/(cx + d) + C/(cx + d)² for a repeated factor.
• Find the constants by substituting the root of each factor, then comparing coefficients for the rest.
• Integrate each term. The squared term gives −1/(p(px + q)), not a logarithm.

Reminder. 5/((x − 1)(x + 4)) = 1/(x − 1) − 1/(x + 4), so the integral is ln|x − 1| − ln|x + 4| + c = ln|(x − 1)/(x + 4)| + c.

Which method?

Integrand looks like	First try
f’(x) times a power of f(x)	Recognition, or u = f(x)
A fraction whose top is a multiple of the derivative of the bottom	k ln
A linear factor times a power of a linear bracket	u = the bracket
A square root of x or of a linear expression	u = the root
√(a² − x²) or 1/√(a² − x²)	x = a sin θ
x^n times eˣ, sin x or cos x	Parts, u = x^n
ln x, or x^n times ln x	Parts, u = ln x
A fraction with two or three linear factors below	Partial fractions

Must-know distinctions

• Recognition vs substitution: recognition is substitution done in your head. Both come from the chain rule.
• ln|f| vs [f]ⁿ⁺¹: a power of −1 gives a logarithm, not a power of 0.
• Parts vs substitution: parts is for a product of unrelated functions (x and eˣ). Substitution is for a function inside a function.
• Upper/lower bound vs under/over-estimate: rectangles give bounds. The trapezium rule gives a single estimate whose sign of error depends on concavity.
• Limits: with substitution, the limits change. With parts, they stay in x.

Quick self-test

1. Find ∫ 8x(x² + 1)³ dx.
2. Find ∫ x/(x² + 9) dx.
3. Find ∫ cos x/sin x dx.
4. Find ∫ x eˣ dx.
5. Find ∫ ln x dx.
6. Evaluate lim (δx→0) Σ (x = 1 to 4) x² δx.
7. Write 6/((x − 1)(x + 2)) in partial fractions and integrate it.
8. Use u = x + 1 to evaluate ∫₀³ x/√(x + 1) dx.
9. Estimate ∫₀¹ eˣ dx using the trapezium rule with 2 strips, to 3 d.p. Is it an over- or underestimate?
10. For an increasing function, which rectangles give a lower bound for the area?
11. Find ∫ 1/(3x + 2) dx.
12. Find ∫ x cos x dx, and show how you chose u.

Answers

1. (x² + 1)⁴ + c
2. ½ ln(x² + 9) + c
3. ln|sin x| + c
4. (x − 1)eˣ + c
5. x ln x − x + c
6. ∫₁⁴ x² dx = (64 − 1)/3 = 21
7. 2/(x − 1) − 2/(x + 2), giving 2 ln|x − 1| − 2 ln|x + 2| + c
8. Integrand becomes (u − 1)u^(−1/2) with u from 1 to 4: [(2/3)u^(3/2) − 2u^(1/2)]₁⁴ = 8/3
9. h = 0.5: 0.25[1 + e + 2e^0.5] ≈ 1.754. The curve is convex, so it is an overestimate (the exact value is e − 1 ≈ 1.718).
10. Left-edge rectangles, because each lies below the curve.
11. (1/3) ln|3x + 2| + c, because dividing by the coefficient of x undoes the chain rule.
12. Take u = x and dv/dx = cos x, because x becomes 1 when differentiated. The answer is x sin x + cos x + c.

Where marks are usually lost

• Forgetting the constant of integration on an indefinite integral.
• Not adjusting the constant when the numerator is a multiple of f’(x), for example dividing by 6 instead of multiplying by 1/6.
• Leaving x in the integrand after a substitution, or keeping the old limits.
• Choosing the wrong part as u, so the integral gets harder instead of simpler.
• Sign errors when integrating by parts twice.
• Dividing by the coefficient of x incorrectly in ∫ 1/(px + q) dx.
• Giving the trapezium estimate without saying whether it is an over- or underestimate, or without a reason.
• Giving an exact answer as a rounded decimal when the question asks for an exact value.

Take a free 10-minute diagnostic to see which method needs more work.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Content statements 1.08g, 1.08h, 1.08i, 1.08j and 1.09f.