Use the substitution \(u = {x^{\frac{1}{2}}}\) to find \(\int {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}} \).

Hence find the value of \(\frac{1}{2}\int\limits_1^9 {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}} \), expressing your answer in the form arctan \(q\), where \(q \in \mathbb{Q}\).

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{2}{x^{ - \frac{1}{2}}}\) (accept \({\text{d}}u = \frac{1}{2}{x^{ - \frac{1}{2}}}{\text{d}}x\) or equivalent)       A1

substitution, leading to an integrand in terms of \(u\)     M1

\(\int {\frac{{2u{\text{d}}u}}{{{u^3} + u}}} \) or equivalent      A1

= 2 arctan \(\left( {\sqrt x } \right)\left( { + c} \right)\)     A1

[4 marks]

\(\frac{1}{2}\int\limits_1^9 {\frac{{{\text{d}}x}}{{{x^{\frac{3}{2}}} + {x^{\frac{1}{2}}}}}} \) = arctan 3 − arctan 1     A1

tan(arctan 3 − arctan 1) = \(\frac{{3 - 1}}{{1 + 3 \times 1}}\)      (M1)

tan(arctan 3 − arctan 1) = \(\frac{1}{2}\)

arctan 3 − arctan 1 = arctan \(\frac{1}{2}\)     A1

[3 marks]