AQA Further Paper 1 2024 June — Question 11 5 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeIntegration by parts with inverse trig
DifficultyStandard +0.3 Part (a) is a straightforward product rule application with a standard derivative. Part (b) requires recognizing the result from (a) to reverse the product rule (integration by parts), which is a standard technique. While this is Further Maths, the question is relatively routine once the connection is spotted, requiring only mechanical application of familiar methods with minimal problem-solving insight.
Spec1.08i Integration by parts4.08g Derivatives: inverse trig and hyperbolic functions
  1. Find \(\frac{d}{dx}(x^2\tan^{-1} x)\) [1 mark]
  2. Hence find \(\int 2x \tan^{-1} x \, dx\) [4 marks]
Question 11:
AnswerMarks
11(a)x2
Obtains 2xtan–1 x+
AnswerMarks Guidance
1+x21.1b B1
(x2tan–1 x) = 2xtan–1 x+
dx 1+x2
AnswerMarks Guidance
Subtotal1
QMarking instructions AO
AnswerMarks
11(b)Uses result of part (a) to
obtain an equation involving
AnswerMarks Guidance
the required integral.2.2a M1
( x 2 t a n – 1 x = ) 2 x t a n – 1 x +
d x 1 + x 2
x 2
  2 x t a n – 1 x d x +  d x = x 2 t a n – 1 x
1 + x 2
x 2 + x 1 2 – 1
 d x =  d x
1 + x 2 + 1 2 x
1
=  d x 1  – d x
1 + x 2
= x – t a – 1 n x + c
x2
2xtan–1 x dx= x2tan–1 x– dx
1+x2
= x2tan–1 x–x+tan–1 x+c
Deduces that
x2 1
=1−
1+x2 1+x2
or
uses the substitution
x = t a n u to obtain
AnswerMarks Guidance
sec2u−12.2a M1
Obtains x – t a n – 1 x1.1b A1
Completes a reasoned
argument starting with the
result from part (a) to obtain
AnswerMarks Guidance
x 2 t a n – 1 x – x + t a n – 1 x + c2.1 R1
Subtotal4
Question total5
QMarking
instructionsAO Marks 
Question 11:
--- 11(a) ---
11(a) | x2
Obtains 2xtan–1 x+
1+x2 | 1.1b | B1 | d x2
(x2tan–1 x) = 2xtan–1 x+
dx 1+x2
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 11(b) ---
11(b) | Uses result of part (a) to
obtain an equation involving
the required integral. | 2.2a | M1 | d x 2
( x 2 t a n – 1 x = ) 2 x t a n – 1 x +
d x 1 + x 2
x 2
  2 x t a n – 1 x d x +  d x = x 2 t a n – 1 x
1 + x 2
x 2 + x 1 2 – 1
 d x =  d x
1 + x 2 + 1 2 x
1
=  d x 1  – d x
1 + x 2
= x – t a – 1 n x + c
x2
2xtan–1 x dx= x2tan–1 x– dx
1+x2
= x2tan–1 x–x+tan–1 x+c
Deduces that
x2 1
=1−
1+x2 1+x2
or
uses the substitution
x = t a n u to obtain
sec2u−1 | 2.2a | M1
Obtains x – t a n – 1 x | 1.1b | A1
Completes a reasoned
argument starting with the
result from part (a) to obtain
x 2 t a n – 1 x – x + t a n – 1 x + c | 2.1 | R1
Subtotal | 4
Question total | 5
Q | Marking
instructions | AO | Marks | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Find $\frac{d}{dx}(x^2\tan^{-1} x)$ [1 mark]

\item Hence find $\int 2x \tan^{-1} x \, dx$ [4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2024 Q11 [5]}}
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