## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $u = 1+x \Rightarrow \int_0^3 x(1+x)^{-1/2}\,dx = \int_1^4 (u-1)u^{-1/2}\,du$ | M1 | $\int(u-1)u^{-1/2}(du)$. Condone no $du$, missing bracket, ignore limits |
| $= \int_1^4 (u^{1/2} - u^{-1/2})\,du$ | A1 | $\int(u^{1/2}-u^{-1/2})(du)$ |
| $= \left[\frac{2}{3}u^{3/2} - 2u^{1/2}\right]_1^4$ | A1 | $\left[\frac{2}{3}u^{3/2} - 2u^{1/2}\right]$ o.e. e.g. $\left[\frac{u^{3/2}}{3/2} - \frac{u^{1/2}}{1/2}\right]$; ignore limits |
| $= (16/3 - 4) - (2/3 - 2)$ | M1dep | Upper–lower dep 1st M1 and integration. With correct limits e.g. 1, 4 for $u$ or 0, 3 for $x$ |
| $= 2\frac{2}{3}$ | A1cao | Or $2.\dot{6}$ but must be exact |
| **OR** Let $u=x$, $v'=(1+x)^{-1/2}$ | M1 | Or using $w=(1+x)^{1/2} \Rightarrow \int\frac{(w^2-1)2w}{w}(dw)$ M1 $= \int 2(w^2-1)(dw)$ A1 $= \left[\frac{2}{3}w^3-2w\right]$ A1 |
| $\Rightarrow u'=1$, $v=2(1+x)^{1/2}$ | A1 | Upper–lower with correct limits ($w=1,2$) M1 |
| $\int_0^3 x(1+x)^{-1/2}\,dx = \left[2x(1+x)^{1/2}\right]_0^3 - \int_0^3 2(1+x)^{1/2}\,dx$ | A1 | Ignore limits, condone no $dx$. $8/3$ A1 cao |
| $= \left[2x(1+x)^{1/2} - \frac{4}{3}(1+x)^{3/2}\right]_0^3$ | A1 | Ignore limits |
| $= (2\times3\times2 - 4\times8/3) - (0-4/3)$ | | |
| $= 2\frac{2}{3}$ | A1cao | Or $2.\dot{6}$ but must be exact |
| **[5]** | | |

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