Using the substitution $u^2 = 2x - 1$, or otherwise, find the exact value of $$\int_{1}^{3} \frac{3x}{\sqrt{2x-1}} \, dx.$$ - Edexcel - A-Level Maths Pure - Question 5 - 2006 - Paper 7

We can now rewrite our integral:
$\int_{1}^{3} \frac{3x}{\sqrt{2x - 1}} \, dx = \int_{1}^{\sqrt{5}} \frac{3 \cdot \frac{u^2 + 1}{2}}{u} \cdot \frac{du}{u} = \frac{3}{2} \int_{1}^{\sqrt{5}} (u + \frac{1}{u}) \, du.$

Now we can integrate:
$\int (u + \frac{1}{u}) \, du = \frac{u^2}{2} + \ln |u| + C.$
Thus, the integral evaluates to:
$\frac{3}{2} \left[ \frac{u^2}{2} + \ln |u| \right]_{1}^{\sqrt{5}}.$

Substituting the limits into our expression gives:
$\frac{3}{2} \left( \frac{5}{2} + \ln |\sqrt{5}| - \left( \frac{1}{2} + \ln |1| \right) \right) = \frac{3}{2} \left( \frac{5}{2} - \frac{1}{2} + \ln |\sqrt{5}| \right) = \frac{3}{2} \left( 2 + \frac{1}{2} \ln(5) \right).$

Finally, simplifying gives us the exact value:
$\frac{3}{2} \cdot 2 + \frac{3}{4} \ln(5) = 3 + \frac{3}{4} \ln(5).$