Challenging +1.2 This is a Further Maths question requiring completion of the square to transform the integral into standard form ∫1/√(u²+a²) du = sinh⁻¹(u/a) or ln form, then evaluating at limits. While it requires multiple steps (completing the square, substitution, applying the standard result, and simplifying logarithms), these are all standard techniques for Further Maths students. The 'show that' format with integer parameters makes it more routine than open-ended problems.
16 In this question you must show detailed reasoning.
Show that \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { \mathrm { x } ^ { 2 } + \mathrm { x } + 1 } } \mathrm { dx } = \ln \left( \frac { \mathrm { a } + \mathrm { b } \sqrt { 3 } } { \mathrm { c } } \right)\), where \(a , b\) and \(c\) are integers to be determined.
Any correct substitution for their integral, e.g. 𝑥+ =
2
√3
sinh𝑢
2
[𝑢]arsinh √3
e.g. , ignore limits. Correct substitution must
1
arsinh
√3
have been made for this mark.
A correct expression in logarithmic form (need not be
simplified)
Combining logs.
www. Can be implied by correct final answer provided
two log terms shown.
Alternative method
1
1
1 1
d x = d x
0 x 2 + x + 1 ( x + 12 ) 2 + 34
Answer
Marks
Guidance
0
B1
Completing the square
1
x+1 arsinh 2 or ln ( x + 12 + ( x + 12 ) 2 + 34 ) 1
Answer
Marks
1 2 3 0 0
M1*
A1
By inspection for their integral, no slips
Correct, ignore limits
( ) 3 =ln 3+2 −ln or ln ( 32 + 3 ) − ln 32
Answer
Marks
Guidance
3
A1
A correct expression in logarithmic form (need not be
simplified)
3+2 3
=ln
Answer
Marks
3
M1dep
A1
Combining logs.
www. Can be implied by correct final answer provided
two log terms shown.
M1*
A1
By inspection for their integral, no slips
Correct, ignore limits
A correct expression in logarithmic form (need not be
simplified)
M1dep
A1
[6]
Question 16:
16 | DR
1
1
1 1
dx= dx
0 x2+x+1 ( x+1)2 +3
0 2 4
let u = x + 12 d u = d x , 3 1 giving 2 du
1 u2+3
2
4
3
= arsinh u 2 or ln ( u+ u2+3 ) 3 2
1 3 4 1
2 1
2
2
1
[= arsinh√3−arsinh ]
√3
=ln ( 3+2 ) −ln 3 or ln ( 3+ 3 ) −ln3
3 2 2
3 + 2 3
= l n
3 | B1
M1*
A1
A1
M1dep
A1 | 3.1a
2.1
2.1
2.1
2.1
2.1 | Completing the square
1
Any correct substitution for their integral, e.g. 𝑥+ =
2
√3
sinh𝑢
2
[𝑢]arsinh √3
e.g. , ignore limits. Correct substitution must
1
arsinh
√3
have been made for this mark.
A correct expression in logarithmic form (need not be
simplified)
Combining logs.
www. Can be implied by correct final answer provided
two log terms shown.
Alternative method
1
1
1 1
d x = d x
0 x 2 + x + 1 ( x + 12 ) 2 + 34
0 | B1 | Completing the square
1
x+1 arsinh 2 or ln ( x + 12 + ( x + 12 ) 2 + 34 ) 1
1 2 3 0 0 | M1*
A1 | By inspection for their integral, no slips
Correct, ignore limits
( ) 3 =ln 3+2 −ln or ln ( 32 + 3 ) − ln 32
3 | A1 | A correct expression in logarithmic form (need not be
simplified)
3+2 3
=ln
3 | M1dep
A1 | Combining logs.
www. Can be implied by correct final answer provided
two log terms shown.
M1*
A1
By inspection for their integral, no slips
Correct, ignore limits
A correct expression in logarithmic form (need not be
simplified)
M1dep
A1
[6]
16 In this question you must show detailed reasoning.
Show that $\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { \mathrm { x } ^ { 2 } + \mathrm { x } + 1 } } \mathrm { dx } = \ln \left( \frac { \mathrm { a } + \mathrm { b } \sqrt { 3 } } { \mathrm { c } } \right)$, where $a , b$ and $c$ are integers to be determined.
\hfill \mbox{\textit{OCR MEI Further Pure Core 2024 Q16 [6]}}