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1.9.27 Integration by substitution (reverse chain rule) (A-level only)

1.9.27 Integration by substitution (reverse chain rule) (A-level only)

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Question 7

Determine the following indefinite integrals:

i.
∫21x2−143x3−6x+5 dx \int \frac{21x^2 - 14}{3x^3 - 6x + 5} \, \mathrm{d}x ∫3x3−6x+521x2−14​dx
[3]
ii.
∫8e4x(7e4x+2)5 dx \int \frac{8\mathrm{e}^{4x}}{(7\mathrm{e}^{4x} + 2)^5} \, \mathrm{d}x ∫(7e4x+2)58e4x​dx
[3]
Markscheme

1.9.27 Integration by substitution (reverse chain rule) (A-level only) Questions

  1. A Level
  2. /Maths
  3. /1.9.27 Integration by substitution (reverse chain rule) (A-level only)

36 exam-style questions on OCR (MEI) A Level Maths 1.9.27 Integration by substitution (reverse chain rule) (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.

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