Integrate the left-hand side of the equation.

$\int_{1}^{a} (\frac{1}{x} - \frac{1}{2 x + 3}) d x = {[\ln x - \frac{1}{2} \ln (2 x + 3)]}_{1}^{a}$

$\ln (x)$ [1]

$- \frac{1}{2} \ln (2 x + 3)$ [1]

Apply the limits.

$(\ln a - \frac{1}{2} \ln (2 a + 3)) - (\ln (1) - \frac{1}{2} \ln (5))$

[1]

Expand the brackets and set equal to the right-hand side.

$\ln a - \frac{1}{2} \ln (2 a + 3) + \frac{1}{2} \ln (5) = \ln 3$

Apply the law of logarithms, $\ln a - \ln b = \ln (\frac{a}{b})$ .

$\ln a + \frac{1}{2} \ln (\frac{5}{2 a + 3}) = \ln 3$

Subtract $\ln a$ from both sides and apply the same law again.

$\frac{1}{2} \ln (\frac{5}{2 a + 3}) = \ln (\frac{3}{a})$

Apply the log law $a \ln b = \ln b^{a}$ .

$\ln [{(\frac{5}{2 a + 3})}^{\frac{1}{2}}] = \ln (\frac{3}{a})$

[1]

Take exponentials of both sides.

${(\frac{5}{2 a + 3})}^{\frac{1}{2}} = \frac{3}{a}$

Square both sides.

$\frac{5}{2 a + 3} = \frac{9}{a^{2}}$

Multiply through by the denominators.

$5 a^{2} = 9 (2 a + 3)$

Expand the brackets and then rearrange to make the equation equal to 0.

$5 a^{2} - 18 a - 27 = 0$

[1]

Use the quadratic formula (and/or a calculator) with $a = 5, b = - 18, c = - 27$ .
Disregard the negative solution since $a > 0$ .

$a = \frac{9 + 6 \sqrt{6}}{5}$ [1]

Integration (Cambridge O Level Additional Maths)

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