2. Use the substitution \(u = 2 ^ { x }\) to find the exact value of
$$\int _ { 0 } ^ { 1 } \frac { 2 ^ { x } } { \left( 2 ^ { x } + 1 \right) ^ { 2 } } d x$$
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Question 2:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\frac{du}{dx} = 2^x \ln 2 \Rightarrow \frac{dx}{du} = \frac{1}{2^x \ln 2}\) B1
\(\frac{du}{dx} = 2^x \ln 2\) or \(\frac{du}{dx} = u\ln 2\) or \(\left(\frac{1}{u}\right)\frac{du}{dx} = \ln 2\)
\(\int \frac{2^x}{(2^x+1)^2}\,dx = \left(\frac{1}{\ln 2}\right)\int \frac{1}{(u+1)^2}\,du\) M1*
\(k\int \frac{1}{(u+1)^2}\,du\) where \(k\) is constant
\((u+1)^{-2} \to a(u+1)^{-1}\) M1
\(= \left(\frac{1}{\ln 2}\right)\left(\frac{-1}{u+1}\right) + c\) A1
\((u+1)^{-2} \to -1\cdot(u+1)^{-1}\)
Change limits: when \(x=0\), \(u=1\); when \(x=1\), \(u=2\) \(= \frac{1}{\ln 2}\left[\frac{-1}{(u+1)}\right]_1^2 = \frac{1}{\ln 2}\left[\left(-\frac{1}{3}\right)-\left(-\frac{1}{2}\right)\right]\) depM1*
Correct use of limits \(u=1\) and \(u=2\)
\(= \frac{1}{6\ln 2}\) A1 aef
\(\frac{1}{6\ln 2}\) or \(\frac{1}{\ln 4}-\frac{1}{\ln 8}\) or \(\frac{1}{2\ln 2}-\frac{1}{3\ln 2}\). Exact value only!
Total: 6 marks [6]
Other acceptable answers for A1: \(\frac{1}{2\ln 8}\) or \(\frac{1}{\ln 64}\). Use calculator to check ≈ 0.240449…
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# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{du}{dx} = 2^x \ln 2 \Rightarrow \frac{dx}{du} = \frac{1}{2^x \ln 2}$ | B1 | $\frac{du}{dx} = 2^x \ln 2$ or $\frac{du}{dx} = u\ln 2$ or $\left(\frac{1}{u}\right)\frac{du}{dx} = \ln 2$ |
| $\int \frac{2^x}{(2^x+1)^2}\,dx = \left(\frac{1}{\ln 2}\right)\int \frac{1}{(u+1)^2}\,du$ | M1* | $k\int \frac{1}{(u+1)^2}\,du$ where $k$ is constant |
| $(u+1)^{-2} \to a(u+1)^{-1}$ | M1 | |
| $= \left(\frac{1}{\ln 2}\right)\left(\frac{-1}{u+1}\right) + c$ | A1 | $(u+1)^{-2} \to -1\cdot(u+1)^{-1}$ |
| Change limits: when $x=0$, $u=1$; when $x=1$, $u=2$ | | |
| $= \frac{1}{\ln 2}\left[\frac{-1}{(u+1)}\right]_1^2 = \frac{1}{\ln 2}\left[\left(-\frac{1}{3}\right)-\left(-\frac{1}{2}\right)\right]$ | depM1* | Correct use of limits $u=1$ and $u=2$ |
| $= \frac{1}{6\ln 2}$ | A1 aef | $\frac{1}{6\ln 2}$ or $\frac{1}{\ln 4}-\frac{1}{\ln 8}$ or $\frac{1}{2\ln 2}-\frac{1}{3\ln 2}$. Exact value only! |
| **Total: 6 marks** | [6] | Other acceptable answers for A1: $\frac{1}{2\ln 8}$ or $\frac{1}{\ln 64}$. Use calculator to check ≈ 0.240449… |
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2. Use the substitution $u = 2 ^ { x }$ to find the exact value of
$$\int _ { 0 } ^ { 1 } \frac { 2 ^ { x } } { \left( 2 ^ { x } + 1 \right) ^ { 2 } } d x$$
\hfill \mbox{\textit{Edexcel C4 2007 Q2 [6]}}