Pre-U Pre-U 9794/1 2010 June — Question 4 5 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
Year2010
SessionJune
Marks5
TopicIntegration by Substitution
TypeSquare root substitution: indefinite integral
DifficultyModerate -0.3 This is a straightforward substitution question with a hint provided. The substitution u = 1 + √x leads directly to du = 1/(2√x) dx, requiring algebraic manipulation to express the integral in terms of u, then integrating and substituting back. While it requires careful handling of the substitution mechanics, it's a standard technique with no conceptual surprises, making it slightly easier than average.
Spec1.08h Integration by substitution

Using the substitution \(u = 1 + \sqrt{x}\), or otherwise, find \(\int \frac{1}{1 + \sqrt{x}} dx\) giving your answer in terms of \(x\). [5]

AnswerMarks Guidance
State or imply a substitution rule for \(x\) and \(dx\) in terms of \(u\)M1
Substitute for \(x\) and \(dx\) in the integral to obtain a final integral in \(u\)M1
Obtain \(2\left[\left(1-\frac{1}{u}\right)du\right]\) or \(2\left[\frac{u-1}{u}\right]du\)A1
Use or imply \(\int\frac{1}{u}du = \ln u\) anywhereB1
Obtain \(2\left[\left(1+\sqrt{x}-\ln\left1+\sqrt{x}\right \right)+c\right]\) OR \(2\left[1+\sqrt{x}+\ln\left
(Do not penalise lack of modulus sign) [5]
OR using parts
AnswerMarks Guidance
State or imply \(u^2 = x\) and \(dx = 2u\,du\)M1
Substitute for \(x\) and \(dx\) in the integral to obtain a final integral in \(u\)M1
Obtain \(2\left[\frac{u}{1+u}\right]du\)A1
Attempt to integrate by parts to obtain at least \(k\left(u + \lnu - \ln 1+u
Obtain \(2\left[\left(1+\sqrt{x}-\ln\left1+\sqrt{x}\right \right)+c\right]\) OR \(2\left[1+\sqrt{x}+\ln\left
(Do not penalise lack of modulus sign) [5]
State or imply a substitution rule for $x$ and $dx$ in terms of $u$ | M1 |
Substitute for $x$ and $dx$ in the integral to obtain a final integral in $u$ | M1 |
Obtain $2\left[\left(1-\frac{1}{u}\right)du\right]$ or $2\left[\frac{u-1}{u}\right]du$ | A1 |
Use or imply $\int\frac{1}{u}du = \ln u$ anywhere | B1 |
Obtain $2\left[\left(1+\sqrt{x}-\ln\left|1+\sqrt{x}\right|\right)+c\right]$ OR $2\left[1+\sqrt{x}+\ln\left|\frac{1}{1+\sqrt{x}}\right|\right]+c$ OR $2\left[\sqrt{x}-\ln\left|1+\sqrt{x}\right|\right]+c$ | A1 |
(Do not penalise lack of modulus sign) | | [5]

OR using parts

State or imply $u^2 = x$ and $dx = 2u\,du$ | M1 |
Substitute for $x$ and $dx$ in the integral to obtain a final integral in $u$ | M1 |
Obtain $2\left[\frac{u}{1+u}\right]du$ | A1 |
Attempt to integrate by parts to obtain at least $k\left(u + \ln|u - \ln|1+u|\right)$ | B1 |
Obtain $2\left[\left(1+\sqrt{x}-\ln\left|1+\sqrt{x}\right|\right)+c\right]$ OR $2\left[1+\sqrt{x}+\ln\left|\frac{1}{1+\sqrt{x}}\right|\right]+c$ OR $2\left[\sqrt{x}-\ln\left|1+\sqrt{x}\right|\right]+c$ | A1 |
(Do not penalise lack of modulus sign) | | [5]
Using the substitution $u = 1 + \sqrt{x}$, or otherwise, find $\int \frac{1}{1 + \sqrt{x}} dx$ giving your answer in terms of $x$. [5]

\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2010 Q4 [5]}}