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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover 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), Pure Mathematics 4 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.

Revise first with the study guide and revision notes. Course pages: Edexcel IAL Mathematics hub and the checklist. Give exact answers unless a question says otherwise.

Questions

1. (a) Find ∫ 5x/(x² + 3) dx. [2] (b) Evaluate ∫ from 1 to 2 of 12/(2x − 1)⁴ dx. [3]

2. Use the substitution u = 5 − x to find ∫ x√(5 − x) dx. [5]

3. Evaluate ∫ from 0 to π/2 of x sin 2x dx. [5]

4. Evaluate ∫ from 1 to 4 of (ln x)/√x dx. [5]

5. Find ∫ x² sin x dx. [5]

6. f(x) = (7x + 3)/((x − 1)(x + 4)).

(a) Write f(x) as partial fractions A/(x − 1) + B/(x + 4), stating the values of A and B. [3] (b) Hence evaluate ∫ from 2 to 6 of f(x) dx, leaving your answer in terms of ln 5 and ln 3. [3]

7. g(x) = (2x² + 2x − 9)/((x + 1)(x − 2)).

(a) Show that g(x) can be written as P + A/(x + 1) + B/(x − 2), and find P, A and B. [4] (b) Hence evaluate ∫ from 3 to 5 of g(x) dx, giving the answer in the form a + ln b. [3]

8. Use the substitution u = 1 + √x to evaluate ∫ from 1 to 9 of 1/(√x (1 + √x)²) dx. [5]

9. Find ∫ e^(−x) sin 2x dx. [6]

10. Use the substitution u = x² to show that ∫ from 0 to 1 of x³ e^(x²) dx = ½. [8]

11. Show that ∫ from 1 to e of (ln x)/x³ dx = (e² − 3)/(4e²). [6]

Answers

1. (a) The numerator is a multiple of the derivative of x² + 3, so the result is a logarithm: k ln(x² + 3) [1]. Differentiating gives 2kx/(x² + 3), so k = 5/2: (5/2) ln(x² + 3) + c [1] (b) Write 12(2x − 1)^(−4). Antiderivative: 12 × (2x − 1)^(−3)/((−3)(2)) [1], which is −2(2x − 1)^(−3) [1]. At x = 2: −2/27. At x = 1: −2. The difference is 2 − 2/27 = 52/27 [1] Examiner insight: In part (a) a modulus is not needed because x² + 3 is positive, and writing the reasoning for the factor 5/2 shows the method even if the constant slips; in part (b) the factor 1/((−3)(2)) is where most errors occur, so write it out.

2. u = 5 − x gives x = 5 − u and dx = −du [1]. ∫ (5 − u)√u (−du) = ∫ (u^(3/2) − 5u^(1/2)) du [1] = (2/5)u^(5/2) [1] − (10/3)u^(3/2) [1] Substituting back: (2/5)(5 − x)^(5/2) − (10/3)(5 − x)^(3/2) + c [1] Examiner insight: The minus sign from dx = −du is easy to drop; stating dx in terms of du on its own line, before rewriting the integrand, protects the whole question.

3. Take u = x and dv/dx = sin 2x, so du/dx = 1 and v = −½ cos 2x [1]. ∫ x sin 2x dx = −(x/2) cos 2x + ∫ ½ cos 2x dx [1] = −(x/2) cos 2x + ¼ sin 2x [1] At x = π/2: −(π/4)(−1) + 0 = π/4 [1]. At x = 0: 0. The value is π/4 [1] Examiner insight: Show the choice of u and v and the evaluation at both limits. If one sign slips, a clear structure shows where the method was sound.

4. Take u = ln x and dv/dx = x^(−1/2), so du/dx = 1/x and v = 2x^(1/2) [1]. ∫ (ln x)/√x dx = 2√x ln x − ∫ 2x^(−1/2) dx [1] = 2√x ln x − 4√x [1] At x = 4: 4 ln 4 − 8 = 8 ln 2 − 8. At x = 1: 0 − 4 [1]. The value is 8 ln 2 − 4 [1] Examiner insight: Choose ln x as u: its derivative 1/x makes the new integral a simple power. Taking u = x^(−1/2) would leave you needing the integral of ln x in the next step, which is harder than the original.

5. First pass: u = x², dv/dx = sin x, so du/dx = 2x and v = −cos x [1]. ∫ x² sin x dx = −x² cos x + ∫ 2x cos x dx [1] Second pass: u = 2x, v = sin x, so ∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx [1] = 2x sin x + 2 cos x [1] Result: −x² cos x + 2x sin x + 2 cos x + c [1] Examiner insight: Keep the first term in brackets when you subtract the second integral, and check by differentiating: the terms should collapse to x² sin x.

6. (a) Set 7x + 3 = A(x + 4) + B(x − 1) [1]. Put x = 1: 10 = 5A, so A = 2 [1]. Put x = −4: −25 = −5B, so B = 5 [1]. (b) ∫ f(x) dx = 2 ln|x − 1| + 5 ln|x + 4| [1]. From 2 to 6: (2 ln 5 + 5 ln 10) − (2 ln 1 + 5 ln 6) [1] = 2 ln 5 + 5 ln(5/3) = 7 ln 5 − 5 ln 3 [1] Examiner insight: A common slip is to leave 5 ln 10 − 5 ln 6 unsimplified, so the answer is not in terms of ln 5 and ln 3 only. Combine the logarithms of 5 first to reach 7 ln 5 − 5 ln 3.

7. (a) The denominator is x² − x − 2, and the numerator has the same degree, so divide first. 2x² + 2x − 9 = 2(x² − x − 2) + (4x − 5), so P = 2 [1] and the remainder is 4x − 5 [1]. Then 4x − 5 = A(x − 2) + B(x + 1) [1]. Put x = 2: 3 = 3B, so B = 1. Put x = −1: −9 = −3A, so A = 3 [1]. So g(x) = 2 + 3/(x + 1) + 1/(x − 2). (b) ∫ g(x) dx = 2x + 3 ln(x + 1) + ln(x − 2) [1] At x = 5: 10 + 3 ln 6 + ln 3. At x = 3: 6 + 3 ln 4 + 0 [1] Difference: 4 + 3 ln(3/2) + ln 3 = 4 + ln(27/8 × 3) = 4 + ln(81/8) [1] Examiner insight: If you apply partial fractions to a top-heavy fraction without dividing first, the constants cannot match, so test the degrees before you start. Check A and B by putting x = 0 into both forms.

8. u = 1 + √x gives du = dx/(2√x), so dx/√x = 2 du [1]. When x = 1, u = 2; when x = 9, u = 4 [1]. The integral becomes ∫ from 2 to 4 of 2/u² du [1] = [−2/u] from 2 to 4 [1] = −½ − (−1) = ½ [1] Examiner insight: When the substitution is given, the work lies in converting dx and the limits accurately. Writing both limits in u removes any need to return to x.

9. First pass: u = sin 2x, dv/dx = e^(−x), so du/dx = 2 cos 2x and v = −e^(−x) [1]. Let I = ∫ e^(−x) sin 2x dx. Then I = −e^(−x) sin 2x + 2 ∫ e^(−x) cos 2x dx [1]. Second pass: u = cos 2x, v = −e^(−x): ∫ e^(−x) cos 2x dx = −e^(−x) cos 2x − 2I [1]. So I = −e^(−x) sin 2x − 2e^(−x) cos 2x − 4I [1], giving 5I = −e^(−x)(sin 2x + 2 cos 2x) [1]. I = −e^(−x)(sin 2x + 2 cos 2x)/5 + c [1] Examiner insight: Use the same type of function as u on both passes. Swapping halfway undoes the first pass, and the integral you started with simply reappears.

10. u = x² gives du = 2x dx, so x³ dx = ½ u du [1]. The limits are u = 0 and u = 1 [1]. The integral becomes ½ ∫ from 0 to 1 of u eᵘ du [1]. Parts with U = u and dV/du = eᵘ: u eᵘ − ∫ eᵘ du [1] = u eᵘ − eᵘ [1]. At u = 1: e − e = 0. At u = 0: −1 [1]. The difference is 0 − (−1) = 1 [1]. Multiplying by ½ gives ½ [1]. Examiner insight: A question that gives you the result to reach rewards a complete chain of reasoning: show the new limits, the integrand in u, the parts step and the evaluation. Agreeing decimals do not replace those lines.

11. Take u = ln x and dv/dx = x^(−3), so du/dx = 1/x and v = −1/(2x²) [1]. ∫ (ln x)/x³ dx = −(ln x)/(2x²) + ∫ 1/(2x³) dx [1] = −(ln x)/(2x²) − 1/(4x²) [1] At x = e: −1/(2e²) − 1/(4e²) = −3/(4e²) [1]. At x = 1: 0 − 1/4 [1]. Difference: −3/(4e²) + 1/4 = (e² − 3)/(4e²) [1] Examiner insight: A common error is to treat 1/x³ as a logarithm. Write it as x^(−3) before integrating, and take care with the sign on the lower limit.

Where marks are usually lost

• Writing dx = du when the substitution has a coefficient, or when it is u = 5 − x, which gives dx = −du.
• Leaving an x in the integrand after substituting, instead of solving for x.
• Applying x-limits to an integral already in u.
• Dropping the sign of the second integral in parts, especially in repeated parts.
• Swapping which function is u between passes in an eˣ with sin or cos integral.
• Integrating a squared denominator as a logarithm, or a first-power denominator as a power.
• Skipping the division step when the numerator’s degree is not lower than the denominator’s.
• Giving an answer as separate logarithms when the question asks for a particular form.
• Omitting + c on an indefinite integral.

Next steps

• Revise any weak method with the revision notes and the study guide.
• Return to the Edexcel IAL Mathematics hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class if you want one-to-one help.

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.