Edexcel C4 — Question 2 6 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks6
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Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeIndefinite integral with non-linear substitution (algebraic/exponential/logarithmic)
DifficultyStandard +0.3 This is a straightforward substitution question where the substitution is explicitly given. Students must find du/dx = -2x, rearrange to get dx in terms of du, substitute, integrate 1/u to get ln|u|, and substitute back. While it requires careful algebraic manipulation and understanding of logarithmic integration, it's a standard C4 technique with no conceptual surprises, making it slightly easier than average.
Spec1.08h Integration by substitution
Use the substitution \(u = 1 - x^2\) to find $$\int \frac{1}{1-x^2} \, dx.$$ [6]
AnswerMarks Guidance
\(u = 1-x^3 \Rightarrow x=(1-u)^{\frac{1}{3}}, \frac{dx}{du} = -2(1-u) = 2u-2\)M1 A1
\(I = \int \frac{1}{u} \times (2u-2) \, du = \int \left(2-\frac{2}{u}\right) du\)A1
\(= 2u - 2\lnu + c\)
\(= 2(1-x^3) - 2\ln1-x^3 + c\) 
$u = 1-x^3 \Rightarrow x=(1-u)^{\frac{1}{3}}, \frac{dx}{du} = -2(1-u) = 2u-2$ | M1 A1 |

$I = \int \frac{1}{u} \times (2u-2) \, du = \int \left(2-\frac{2}{u}\right) du$ | A1 |

$= 2u - 2\ln|u| + c$ | M1 A1 |

$= 2(1-x^3) - 2\ln|1-x^3| + c$ | A1 | **(6 marks)**

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Use the substitution $u = 1 - x^2$ to find
$$\int \frac{1}{1-x^2} \, dx.$$ [6]

\hfill \mbox{\textit{Edexcel C4  Q2 [6]}}
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