C4 January 2009 Q6
6.
\[\int \dfrac{\mathrm{e}^{3x}}{1 + \mathrm{e}^x}\,\mathrm{d}x = \dfrac{1}{2}\mathrm{e}^{2x} - \mathrm{e}^x + \ln(1 + \mathrm{e}^x) + k,\]where \(k\) is a constant. (7)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \tan^2 x\,\mathrm{d}x\) | |
| \(\left[NB\colon\ \underline{\sec^2 A = 1 + \tan^2 A} \text{ gives } \underline{\tan^2 A = \sec^2 A - 1}\right]\) | M1 oe |
| \(= \displaystyle\int \sec^2 x - 1\,\mathrm{d}x\) | |
| \(= \underline{\tan x - x}\ (+\,c)\) | A1 |
| (2) |
Notes
M1 oe: The correct underlined identity.
A1: Correct integration with/without + c
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \dfrac{1}{x^3}\ln x\,\mathrm{d}x\) | |
| \(\left\{\begin{aligned} u &= \ln x &&\Rightarrow\ \tfrac{\mathrm{d}u}{\mathrm{d}x} = \tfrac{1}{x}\\ \tfrac{\mathrm{d}v}{\mathrm{d}x} &= x^{-3} &&\Rightarrow\ v = \tfrac{x^{-2}}{-2} = \tfrac{-1}{2x^2}\end{aligned}\right\}\) | |
| \(= -\dfrac{1}{2x^2}\ln x - \displaystyle\int -\dfrac{1}{2x^2}.\dfrac{1}{x}\,\mathrm{d}x\) | M1 A1 |
| \(= -\dfrac{1}{2x^2}\ln x + \dfrac{1}{2}\displaystyle\int \dfrac{1}{x^3}\,\mathrm{d}x\) | |
| \(= \underline{-\dfrac{1}{2x^2}\ln x + \dfrac{1}{2}\left(-\dfrac{1}{2x^2}\right)}\ (+\,c)\) | M1 A1 oe |
| (4) |
Notes
M1: Use of ‘integration by parts’ formula in the correct direction. A1: Correct expression.
Correct direction means that \(u = \ln x\).
M1: An attempt to multiply through \(\dfrac{k}{x^n},\ n \in \mathbb{Z},\ n \geqslant 2\) by \(\tfrac{1}{x}\) and an attempt to … “integrate” (process the result);
A1 oe: correct solution with/without + c
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \dfrac{\mathrm{e}^{3x}}{1 + \mathrm{e}^x}\,\mathrm{d}x\) | |
| \(\left\{u = 1 + \mathrm{e}^x \Rightarrow \underline{\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x},\ \underline{\dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{\mathrm{e}^x}},\ \underline{\dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{u - 1}}\right\}\) | B1 |
| \(= \displaystyle\int \dfrac{\mathrm{e}^{2x}.\mathrm{e}^x}{1 + \mathrm{e}^x}\,\mathrm{d}x = \int \dfrac{(u - 1)^2.\mathrm{e}^x}{u}.\dfrac{1}{\mathrm{e}^x}\,\mathrm{d}u\) or \(= \displaystyle\int \dfrac{(u - 1)^3}{u}.\dfrac{1}{(u - 1)}\,\mathrm{d}u\) | M1* |
| \(= \displaystyle\int \dfrac{(u - 1)^2}{u}\,\mathrm{d}u\) | A1 |
| \(= \displaystyle\int \dfrac{u^2 - 2u + 1}{u}\,\mathrm{d}u\) | |
| \(= \displaystyle\int u - 2 + \dfrac{1}{u}\,\mathrm{d}u\) | dM1* |
| \(= \dfrac{u^2}{2} - 2u + \ln u\ (+\,c)\) | A1 |
| \(= \dfrac{(1 + \mathrm{e}^x)^2}{2} - 2(1 + \mathrm{e}^x) + \ln(1 + \mathrm{e}^x) + c\) | dM1* |
| \(= \tfrac{1}{2} + \mathrm{e}^x + \tfrac{1}{2}\mathrm{e}^{2x} - 2 - 2\mathrm{e}^x + \ln(1 + \mathrm{e}^x) + c\) | |
| \(= \tfrac{1}{2} + \mathrm{e}^x + \tfrac{1}{2}\mathrm{e}^{2x} - 2 - 2\mathrm{e}^x + \ln(1 + \mathrm{e}^x) + c\) | |
| \(= \tfrac{1}{2}\mathrm{e}^{2x} - \mathrm{e}^x + \ln(1 + \mathrm{e}^x) - \tfrac{3}{2} + c\) | |
| \(= \tfrac{1}{2}\mathrm{e}^{2x} - \mathrm{e}^x + \ln(1 + \mathrm{e}^x) + k\) AG | A1 cso |
| (7) | |
| (13 marks) |
Notes
B1: Differentiating to find any one of the three underlined
M1*: Attempt to substitute for \(\mathrm{e}^{2x} = \mathrm{f}(u)\), their \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{\mathrm{e}^x}\) and \(u = 1 + \mathrm{e}^x\) or \(\mathrm{e}^{3x} = \mathrm{f}(u)\), their \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{1}{u - 1}\) and \(u = 1 + \mathrm{e}^x\).
A1: \(\underline{\displaystyle\int \dfrac{(u - 1)^2}{u}\,\mathrm{d}u}\)
dM1*: An attempt to multiply out their numerator to give at least three terms and divide through each term by \(u\)
A1: Correct integration with/without +c
dM1*: Substitutes \(u = 1 + \mathrm{e}^x\) back into their integrated expression with at least two terms.
A1 cso: \(\tfrac{1}{2}\mathrm{e}^{2x} - \mathrm{e}^x + \ln(1 + \mathrm{e}^x) + k\) must use a \(+\,c\) and “\(-\tfrac{3}{2}\)” combined.