## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + 4x + 8 = (x+2)^2 + 4$ | M1 | Complete the square in order to use standard form |
| $\int_0^1 \frac{1}{\sqrt{x^2+4x+8}}\,dx = \int_0^1 \frac{1}{\sqrt{(x+2)^2+4}}\,dx$ | A1 | |
| | M1 | Use correct standard form in integration |
| $= \left[\sinh^{-1}\frac{x+2}{2}\right]_0^1 = \sinh^{-1}\left(\frac{3}{2}\right) - \sinh^{-1}1$ | A1 | Answer in $\sinh^{-1}$ form |
| $= \ln\left(\frac{3}{2}+\sqrt{1+\frac{9}{4}}\right) - \ln(1+\sqrt{2}) = \ln\left(\frac{3}{2}+\sqrt{\frac{13}{4}}\right) - \ln(1+\sqrt{2})$ | M1 | Attempt to turn into log form |
| $= \ln\left(\frac{3+\sqrt{13}}{2+2\sqrt{2}}\right)$ | A1 | www isw |

**Alternative (log form directly):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \frac{1}{\sqrt{(x+2)^2+4}}\,dx = \left[\ln\left((x+2)+\sqrt{(x+2)^2+4}\right)\right]_0^1$ | M1, A1 | Attempt to use standard form; indefinite integral |
| $= \ln(3+\sqrt{13}) - \ln(2+\sqrt{8}) = \ln\left(\frac{3+\sqrt{13}}{2+2\sqrt{2}}\right)$ | M1, A1 | Limits; www isw |

**Alternative (substitution $x+2 = 2\tan\theta$):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x+2 = 2\tan\theta \Rightarrow I = \left[\ln(\sec\theta+\tan\theta)\right]_{\pi/4}^{\tan^{-1}3/2}$ | M1, A1 | Substitution; indefinite integral |
| $= \ln\left(\frac{3}{2}+\frac{\sqrt{13}}{2}\right) - \ln(1+\sqrt{2}) = \ln\left(\frac{3+\sqrt{13}}{2+2\sqrt{2}}\right)$ | M1, A1 | Deal with limits; www isw |

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