Let $u = 1 + x \Rightarrow \int_0^1 x(1+x)^{-1/2}dx = \int_1^4 (u-1)u^{-1/2}du$ | M1, A1, A1, M1dep [5] | $\int (u-1)u^{-1/2}(du)$ *; condone no $du$, missing bracket, ignore limits; $\int(u^{1/2} - u^{-1/2})(du)$; upper–lower dep 1st M1 and integration; with correct limits e.g. 1, 4 for $u$ or 0, 3 for $x$ or using $w = (1+x)^{1/2} \Rightarrow \int \frac{(w^2-1)2w}{w}(dw)$ M1

$= \int(u^{1/2} - u^{-1/2})du$ |  |  

$= \left[\frac{2}{3}u^{3/2} - 2u^{1/2}\right]_1^4$ |  |  

$= (16/3 - 4) - (2/3 - 2)$ | A1cao | $\left[\frac{2}{3}u^{3/2} - 2u^{1/2}\right]$ o.e.; ignore limits

$= 2\frac{2}{3}$ |  |  

**OR** Let $u = x, v' = (1+x)^{-1/2}$ | M1, A1 | ignore limits, condone no $dx$

$\Rightarrow u' = 1, v = 2(1+x)^{1/2}$ |  |  

$\Rightarrow \int_0^x(1+x)^{-1/2}dx = [2x(1+x)^{1/2}]_0^1 - \int_0^1 2(1+x)^{1/2}dx$ | A1, A1 | ignore limits, condone no $dx$ allow $\int 1u \cdot (-u)$ done by parts: $2u^{1/2}(u-1) - \int 2u^{1/2}du$ A1 $[2u^{1/2}(u-1) - 4u^{3/2}/3]$ A1 substituting correct limits M1 $8/3$ A1cao

$= [2x(1+x)^{1/2} - \frac{4}{3}(1+x)^{3/2}]_0^1$ |  |  

$= (2 \times 3 \times 2 - 4 \times 8/3) - (0 - 4/3)$ |  |  

$= 2\frac{2}{3}$ |  |  

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