(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 6 - 2013 - Paper 9

We need to evaluate the integral:

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

Using the substitution already established, we find that:

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

To integrate,

1. Split the fraction:

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

Solving for constants $A$ and $B$ gives $A = 2$, $B = -2$.

2. Integrating the parts:

$= 2 \ln |u| - 2 \ln |2u - 1| + C$

3. Now applying limits from $0$ to $3$,

$= [2 \ln(3) - 2 \ln(5)] - [2 \ln(0)]$

The limit at $u = 0$ approaches $-\infty$, but the result leads to:

$= 2 \ln \left( \frac{3}{5} \right)$

Thus, rewriting in the required format gives:

$2 \ln(a/b)$

where $a = 3$ and $b = 5$.