let $u = \ln x$, $du/dx = x^3$, $du/dx = 1/x$, $v = \frac{1}{4}x^4$ | M1 | $u, u', v, v'$ all correct
$\int_1^2 x^3 \ln x \, dx = \left[\frac{1}{4}x^4\ln x\right]_1^2 - \int_1^2 \frac{1}{4}x^4 \cdot \frac{1}{x} dx$ | A1 | $\frac{1}{4}x^4 \ln x - \int \frac{1}{4}x^4 \cdot \frac{1}{x}[dx]$
$= \left[\frac{1}{4}x^4\ln x\right]_1^2 - \int_1^2 \frac{1}{4}x^3 dx$ | M1 dep | simplifying $x^4/x = x^3$ in second term (soi)
$= \left[\frac{1}{4}x^4\ln x - \frac{1}{16}x^4\right]_1^2$ | A1 cao | $\frac{1}{4}x^4\ln x - \frac{1}{16}x^4$ o.e.
$= 4\ln 2 - 15/16$ | A1 cao | o.e. must be exact, but can isw

**Total: [5]** | | ignore limits; dep 1st M1; must evaluate ln 1 = 0 and combine –1 + 1/16

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