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 \(u\).