By using the substitution $u = 1 + \sqrt x$, find $\int {\frac{{\sqrt x }}{{1 + \sqrt x }}{\text{d}}x}$.

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{{2\sqrt x }}$     A1

${\text{d}}x = 2(u - 1){\text{d}}u$

Note:     Award the A1 for any correct relationship between ${\text{d}}x$ and ${\text{d}}u$.

$\int {\frac{{\sqrt x }}{{1 + \sqrt x }}{\text{d}}x}  = 2\int {\frac{{{{(u - 1)}^2}}}{u}{\text{d}}u}$     (M1)A1

Note:     Award the M1 for an attempt at substitution resulting in an integral only involving $u$ .

$= 2\int {u - 2 + \frac{1}{u}{\text{d}}u}$     (A1)

$= {u^2} - 4u + 2\ln u( + C)$     A1

$= x - 2\sqrt x  - 3 + 2\ln \left( {1 + \sqrt x } \right)( + C)$     A1

Note:     Award the A1 for a correct expression in $x$, but not necessarily fully expanded/simplified.

[6 marks]

Many candidates worked through this question successfully. A significant minority either made algebraic mistakes with the substitution or tried to work with an integral involving both $x$ and $u$.