Standard +0.3 This is a straightforward application of a given substitution with clear guidance. Students must find du/dx = 1/x, rewrite ln x = u-1, and integrate (u-1)/u² which simplifies to standard forms. While it requires careful algebraic manipulation, the substitution is provided and the technique is routine for C4 level, making it slightly easier than average.
An attempt - not necessarily accurate. No evidence of \(x\) at this stage
Substitute, changing given integral to \(\int \frac{u-1}{u^2}(du)\)
A1
Provided of form \(\frac{au+b}{u^2}\), either split as \(\frac{au}{u^2}+\frac{b}{u^2}\)
M1
or use 'parts' with \(u' = au+b\), \(dv' = \frac{1}{u^2}\)
Integrate as \(\ln u + \frac{1}{u}\) or FT as \(a\ln u - \frac{b}{u}\) \([=F(u)]\)
\(\sqrt{}\)A1
or \(-(au+b)\frac{1}{u}+a\ln u\) FT \([=G(u)]\)
Re-substitute \(u = 1 + \ln x\) in \(F(u)\)
M1
Re-substitute \(u = 1 + \ln x\) in \(G(u)\)
\(\ln(1+\ln x)+\frac{1}{1+\ln x}\) \((+c)\) ISW
A1
or \(\ln(1+\ln x)-\frac{\ln x}{1+\ln x}\) \((+c)\) ISW
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Find $\frac{du}{dx}$ or $\frac{dx}{du}$ | M1 | An attempt - not necessarily accurate. No evidence of $x$ at this stage |
| Substitute, changing given integral to $\int \frac{u-1}{u^2}(du)$ | A1 | |
| Provided of form $\frac{au+b}{u^2}$, either split as $\frac{au}{u^2}+\frac{b}{u^2}$ | M1 | or use 'parts' with $u' = au+b$, $dv' = \frac{1}{u^2}$ |
| Integrate as $\ln u + \frac{1}{u}$ or FT as $a\ln u - \frac{b}{u}$ $[=F(u)]$ | $\sqrt{}$A1 | or $-(au+b)\frac{1}{u}+a\ln u$ FT $[=G(u)]$ |
| Re-substitute $u = 1 + \ln x$ in $F(u)$ | M1 | Re-substitute $u = 1 + \ln x$ in $G(u)$ |
| $\ln(1+\ln x)+\frac{1}{1+\ln x}$ $(+c)$ ISW | A1 | or $\ln(1+\ln x)-\frac{\ln x}{1+\ln x}$ $(+c)$ ISW |
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