# Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = \ln 3x$ and $\frac{dv}{dx} = x^8$ | M1 | Integration by parts as far as $f(x) +/- \int g(x)\,dx$; $x^8$ must be integrated so look for term in $x^9$ |
| $\frac{d}{dx}(\ln 3x) = \frac{1}{x}$ or $\frac{3}{3x}$ | B1 | Stated or clearly used |
| $\frac{x^9}{9}\ln 3x - \int \frac{x^9}{9} \cdot \frac{du}{dx}\,dx$ FT | $\surd$A1 | Correct understanding of 'by parts'; even if $\ln(3x)$ incorrectly differentiated |
| Indication that $\int kx^8\,dx$ is required | M1 | Product of terms must be taken; may already be indicated on previous line |
| $\frac{x^9}{9}\ln 3x - \frac{x^9}{81}$ or $\frac{1}{9}x^9\!\left(\ln 3x - \frac{1}{9}\right)$ ISW $(+c)$ | A1 | $\frac{1}{9}\cdot\frac{x^9}{9}$ to simplify to $\frac{x^9}{81}$; $\frac{3x^9}{243}$ satisfies |

**Alternative (manipulating $\ln(3x)$ first):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln(3x) = \ln 3 + \ln x$ | B1 | |
| $u = \ln x$ and $dv = x^8$ | M1 | In order to find $\int x^8 \ln x\,dx$ |
| $\frac{x^9}{9}\ln x - \int \frac{x^9}{9} \cdot \frac{1}{x}\,dx$ or better | A1 | |
| $\frac{x^9}{9}\ln x - \frac{x^9}{81}$ | A1 | |
| Their $\int x^8 \ln x\,dx + \frac{x^9}{9}\ln 3$ $(+c)$ FT ISW | $\surd$A1 | |

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