(a) Use the substitution $x = u^2$, $u > 0$, to show that \[\int \frac{1}{x(2\sqrt{x} - 1)} dx = \int \frac{2}{u(2u - 1)} du\] (b) Hence show that \[\int_{0}^{9} \frac{1}{x(2\sqrt{x} - 1)} dx = 2 \ln \left( \frac{a}{b} \right)\] where $a$ and $b$ are integers to be determined. - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 9

Using the result from part (a), we can evaluate the integral:

$\int_{0}^{9} \frac{1}{x(2\sqrt{x} - 1)} dx = \int_{0}^{3} \frac{2}{u(2u - 1)} du$

We can simplify this:

1. The integral can be evaluated using partial fractions: $\frac{2}{u(2u - 1)} = \frac{A}{u} + \frac{B}{2u - 1}$

Solving for A and B yields:

• $A = 2$, $B = -2$.

2. Thus: $\int \left( \frac{2}{u} - \frac{2}{2u - 1} \right) du$

Evaluating gives: $2 \ln|u| - 2 \ln|2u - 1|$ Evaluating from 0 to 3 results in: $\left[ 2 \ln(3) - 2 \ln(5) \right]$

This can be simplified to: $2 \ln \left( \frac{3}{5} \right)$

Thus, identifying $a = 3$ and $b = 5$, we conclude: $\int_{0}^{9} \frac{1}{x(2\sqrt{x} - 1)} dx = 2 \ln \left( \frac{a}{b} \right)$