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1.9.27 Integration by substitution (reverse chain rule) (A-level only)
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Question 27
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Markscheme
Determine the following indefinite integrals:
i.
∫21x2−143x3−6x+5 dx \int \frac{21x^2 - 14}{3x^3 - 6x + 5} \, \mathrm{d}x
∫
3
x
3
−
6
x
+
5
21
x
2
−
14
d
x
[3]
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3
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ii.
∫8e4x(7e4x+2)5 dx \int \frac{8\mathrm{e}^{4x}}{(7\mathrm{e}^{4x} + 2)^5} \, \mathrm{d}x
∫
(
7
e
4
x
+
2
)
5
8
e
4
x
d
x
[3]
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Markscheme
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3
Grade answer
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1.9.27 Integration by substitution (reverse chain rule) (A-level only) Questions
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1.9.27 Integration by substitution (reverse chain rule) (A-level only)
