2022 H2 Maths Paper 2 Q1 — Worked Solution

Solution

4 marks

Using the substitution $u = \sqrt{x + 2}$ , so $u^{2} = x + 2$ , giving $x = u^{2} - 2$ and $dx = 2u\, du$ ,

\int\frac{x}{\sqrt{x + 2}}\, dx

= \int\frac{u^{2} - 2}{u} \times 2u\, du

= \int 2\left( u^{2} - 2 \right)\, du

= 2\left( \frac{u^{3}}{3} - 2u \right) + c

= \frac{2u^{3}}{3} - 4u + c

Substituting back $u = \sqrt{x + 2}$ ,

\int\frac{x}{\sqrt{x + 2}}\, dx = \frac{2}{3}(x + 2)^{3/2} - 4\sqrt{x + 2} + c

Factoring out $\sqrt{x + 2}$ ,

= \sqrt{x + 2}\left\lbrack \frac{2}{3}(x + 2) - 4 \right\rbrack + c

= \sqrt{x + 2} \times \frac{2(x + 2) - 12}{3} + c

= \sqrt{x + 2} \times \frac{2x - 8}{3} + c

= \frac{2}{3}(x - 4)\sqrt{x + 2} + c

To check your answer: differentiating

\frac{2}{3}(x - 4)\sqrt{x + 2}

by the product rule gives

\frac{2}{3}\sqrt{x + 2} + \frac{2}{3}(x - 4) \times \frac{1}{2\sqrt{x + 2}} = \frac{2(x + 2) + (x - 4)}{3\sqrt{x + 2}} = \frac{3x}{3\sqrt{x + 2}} = \frac{x}{\sqrt{x + 2}}

✓

$\therefore$ $\int\frac{x}{\sqrt{x + 2}}\, dx = \frac{2}{3}(x - 4)\sqrt{x + 2} + c$ .