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 7 - 2013 - Paper 9

For part (b), we want to calculate:

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

With the substitution $x = u^2$, the limits change as follows:

• When $x = 0$, $u = 0$
• When $x = 9$, $u = 3$

Now we can rewrite the integral as:

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

We can split this into partial fractions:

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

By applying algebra, we find that $A = 2$ and $B = -2$. Thus:

$\int_0^3 \left( \frac{2}{u} - \frac{2}{2u-1} \right) \, du = 2 \ln |u| - 2 \ln |2u-1| \bigg|_0^3.$

Evaluating this from $0$ to $3$:

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

Thus, letting $a = 3$ and $b = 5$, we confirm the result:

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