Edexcel PURE 2024 October — Question 6

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicIntegration by Substitution
TypeIndefinite integral with non-linear substitution (algebraic/exponential/logarithmic)
DifficultyChallenging +1.2 This is a guided substitution question where the substitution is provided and students must execute it systematically. It requires finding du/dx, rearranging to express x² dx in terms of u, then integrating u² terms and back-substituting. While it involves multiple algebraic steps and careful manipulation, the pathway is clear and it's a standard Further Maths Pure technique with no novel insight required. Slightly above average difficulty due to the algebraic complexity and being Further Maths content.
Spec1.08h Integration by substitution
  1. Use the substitution \(u = \sqrt { x ^ { 3 } + 1 }\) to show that
$$\int \frac { 9 x ^ { 5 } } { \sqrt { x ^ { 3 } + 1 } } \mathrm {~d} x = 2 \left( x ^ { 3 } + 1 \right) ^ { k } \left( x ^ { 3 } - A \right) + c$$ where \(k\) and \(A\) are constants to be found and \(c\) is an arbitrary constant.
Question 6:
AnswerMarks
6d u
u = x 3 + 1  u 2 = x 3 + 1  2 u = 3 x 2
d x
or
d x
u = x 3 + 1  u 2 = x 3 + 1  2 u = 3 x 2
d u
or
u = x 3 + 1  d u = 1 ( x 3 + 1 ) − 12  3 x 2
d x 2
or
13 dx 2 − 2
x = ( u 2 − 1 )  = u(u2 −1) 3
AnswerMarks
du 3B1
 9x5 9x5 2u  ( )
e.g. dx= du = 6x3du =6 u2 −1 du
x3+1 u 3x2
or
e.g.  9 x 5 d x =  9 x 5 2 x 3 + 1 d u =  6 x 3 d u = 6  ( u 2 − 1 ) d u
AnswerMarks
x 3 + 1 x 3 + 1 3 x 2M1A1
32 12
 ( ) ( ) ( )
AnswerMarks
6 u 2 − 1 d u = 2 u 3 − 6 u ( + c ) = 2 x 3 + 1 − 6 x 3 + 1 ( + c )M1
12 12
AnswerMarks
= 2 ( x 3 + 1 )  ( x 3 + 1 ) − 3  + c = 2 ( x 3 + 1 ) ( x 3 − 2 ) + cA1
(5)

Total 5

Question 6:
6 | d u
u = x 3 + 1  u 2 = x 3 + 1  2 u = 3 x 2
d x
or
d x
u = x 3 + 1  u 2 = x 3 + 1  2 u = 3 x 2
d u
or
u = x 3 + 1  d u = 1 ( x 3 + 1 ) − 12  3 x 2
d x 2
or
13 dx 2 − 2
x = ( u 2 − 1 )  = u(u2 −1) 3
du 3 | B1
 9x5 9x5 2u  ( )
e.g. dx= du = 6x3du =6 u2 −1 du
x3+1 u 3x2
or
e.g.  9 x 5 d x =  9 x 5 2 x 3 + 1 d u =  6 x 3 d u = 6  ( u 2 − 1 ) d u
x 3 + 1 x 3 + 1 3 x 2 | M1A1
32 12
 ( ) ( ) ( )
6 u 2 − 1 d u = 2 u 3 − 6 u ( + c ) = 2 x 3 + 1 − 6 x 3 + 1 ( + c ) | M1
12 12
= 2 ( x 3 + 1 )  ( x 3 + 1 ) − 3  + c = 2 ( x 3 + 1 ) ( x 3 − 2 ) + c | A1
(5)
Total 5
\begin{enumerate}
  \item Use the substitution $u = \sqrt { x ^ { 3 } + 1 }$ to show that
\end{enumerate}

$$\int \frac { 9 x ^ { 5 } } { \sqrt { x ^ { 3 } + 1 } } \mathrm {~d} x = 2 \left( x ^ { 3 } + 1 \right) ^ { k } \left( x ^ { 3 } - A \right) + c$$

where $k$ and $A$ are constants to be found and $c$ is an arbitrary constant.

\hfill \mbox{\textit{Edexcel PURE 2024 Q6}}
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