a) State or imply $dx = 2t{\text{ }}dt$ or equivalent

Express the integral in terms of $x$ and $dx$

Obtain given answer $\int\limits_1^5 {\left( {2x - 2} \right)ln{\text{ }}x{\text{ }}dx} $, including change of limits

b) Attempt integration by parts obtaining $\left( {\alpha {x^2} + bx} \right)ln{\text{ }}x \pm \int {\left( {\alpha {x^2} + bx} \right)\frac{1}{x}dx} $ or equivalent

Obtain $\left( {{x^2} - 2x} \right)ln{\text{ }}x - \int {\left( {{x^2} - 2x} \right)\frac{1}{x}dx} {\text{ }}$ or equivalent

Obtain $\left( {{x^2} - 2x} \right)ln{\text{ }}x - \frac{1}{2}{x^2} + 2x$

Use limits correctly having integrated twice

Obtain $15{\text{ }}ln{\text{ }}5 - 4$ or exact equivalent

[Equivalent for M1 is $\left( {2x - 2} \right)\left( {\alpha x{\text{ }}ln{\text{ }}x + bx} \right) - \int {\left( {\alpha x{\text{ }}ln{\text{ }}x + bx} \right)2dx} $]