Show that $\int_1^2 {{x^3}\ln x{\text{d}}x = 4\ln 2 - \frac{{15}}{{16}}}$.

any attempt at integration by parts     M1

$u = \ln x \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}$     (A1)

$\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^3} \Rightarrow v = \frac{{{x^4}}}{4}$     (A1)

$= \left[ {\frac{{{x^4}}}{4}\ln x} \right]_1^2 - \int_1^2 {\frac{{{x^3}}}{4}{\text{d}}x}$     A1

Note:     Condone absence of limits at this stage.

$= \left[ {\frac{{{x^4}}}{4}\ln x} \right]_1^2 - \left[ {\frac{{{x^4}}}{{16}}} \right]_1^2$     A1

Note:     Condone absence of limits at this stage.

$= 4\ln 2 - \left( {1 - \frac{1}{{16}}} \right)$     A1

$= 4\ln 2 - \frac{{15}}{{16}}$     AG

[6 marks]