5. (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 5 - 2013 - Paper 9

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

$\int_0^9 \frac{1}{x(2\sqrt{x} - 1)} \, dx = \int_0^3 \frac{2}{u(2u - 1)} \, du$

To evaluate this integral, we break it down using partial fractions:

$\frac{2}{u(2u - 1)} = \frac{A}{u} + \frac{B}{2u - 1}$

Solving for $A$ and $B$, we find: $A = 2, \quad B = -2$

Thus,

$\int \left( \frac{2}{u} - \frac{2}{2u - 1} \right) \, du$

This leads us to:

$2 \ln |u| - 2 \ln |2u - 1|$

Evaluating the limits from $u=0$ to $u=3$ gives: $\left[ 2\ln(3) - 2\ln(5) \right] = 2\ln\left(\frac{3}{5}\right)$

This shows that: $\int_0^9 \frac{1}{x(2\sqrt{x} - 1)} \, dx = 2\ln\left(\frac{3}{5}\right)$

Thus, we can identify $a = 3$ and $b = 5$.