Using the substitution $u = 2 + \sqrt{(2x + 1)}$, or other suitable substitutions, find the exact value of \[ \int_{0}^{1} \frac{1}{2 + \sqrt{(2x + 1)}}\,dx \] giving your answer in the form $A + 2 \ln B$, where $A$ is an integer and $B$ is a positive constant. - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 1

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = \frac{1}{2 \sqrt{(2x + 1)}} \cdot 2 \Rightarrow \frac{du}{dx} = \frac{1}{\sqrt{(2x + 1)}}$

Therefore,

$dx = \sqrt{(2x + 1)} \, du$

When $x = 0$,

$u = 2 + \sqrt{(2 \cdot 0 + 1)} = 2 + 1 = 3$

When $x = 1$,

$u = 2 + \sqrt{(2 \cdot 1 + 1)} = 2 + \sqrt{3}$

The integral becomes:

$\int_{3}^{2 + \sqrt{3}} \frac{1}{2 + (u - 2)} \cdot \sqrt{(2x + 1)} \, du$ Substituting back, we need to find $\sqrt{(2x + 1)}$ in terms of $u$:

$\sqrt{(2x + 1)} = u - 2$ Therefore,

$\int_{3}^{2 + \sqrt{3}} \frac{1}{u} (u - 2) \, du$

This simplifies to:

$\int_{3}^{2 + \sqrt{3}} \left(1 - \frac{2}{u}\right) \, du = \left[u - 2 \ln u \right]_{3}^{2 + \sqrt{3}}$

Evaluating this gives:

$(2 + \sqrt{3} - 2 \ln(2 + \sqrt{3})) - (3 - 2 \ln 3)$

Simplifying this results in:

$2 + 2 \ln 3 - 2 \ln(2 + \sqrt{3})$

This can be expressed as:

$2 + 2 \ln \left( \frac{3}{2 + \sqrt{3}} \right)$

Finally, identifying $A = 2$ and $B = \frac{3}{2 + \sqrt{3}}$, we have:

$A + 2 \ln B$