Solve the inequality $\dfrac{2}{4(x+1)^2+1} > 1$. Hence find $\displaystyle \int_{-1}^{\frac{\sqrt{3}-2}{2}} \left|1 - \dfrac{2}{4(x+1)^2+1}\right|\,dx$, leaving your answer in exact form.

1. Solve the rational inequality by clearing the denominator.

Since $4(x+1)^2 + 1 \geq 1 > 0$ for all real $x$, we can multiply both sides by $4(x+1)^2 + 1$ without changing the inequality sign:
$2 > 4(x+1)^2 + 1$
$1 > 4(x^2 + 2x + 1)$
$4x^2 + 8x + 3 < 0$
$(2x+1)(2x+3) < 0$
So the solution is $-\frac{3}{2} < x < -\frac{1}{2}$.

2. Deduce the sign of the expression inside the absolute value for the definite integral.

From the result above, $\frac{2}{4(x+1)^2+1} > 1 \implies 1 - \frac{2}{4(x+1)^2+1} < 0$ specifically when $-\frac{3}{2} < x < -\frac{1}{2}$.

Thus, over the integral limits:
For $x \in \left[-1, -\frac{1}{2}\right)$, $1 - \frac{2}{4(x+1)^2+1} < 0$.
For $x \in \left[-\frac{1}{2}, \frac{\sqrt{3}-2}{2}\right]$, $1 - \frac{2}{4(x+1)^2+1} \geq 0$.

Hence, the integral must be split at $x = -\frac{1}{2}$.

3. Find the indefinite integral of the rational function.

$\int \frac{2}{4(x+1)^2+1} \,dx = \int \frac{2}{(2x+2)^2+1} \,dx$
By applying the standard integral formula for $\frac{1}{u^2+a^2}$, we obtain:
$= \arctan(2x+2) + C$

4. Split and evaluate the definite integral for the first interval.

$\int_{-1}^{-\frac{1}{2}} \left( \frac{2}{4(x+1)^2+1} - 1 \right)\,dx$
$= \Big[ \arctan(2x+2) - x \Big]_{-1}^{-\frac{1}{2}}$
$= \left( \arctan(1) - \left(-\frac{1}{2}\right) \right) - \Big( \arctan(0) - (-1) \Big)$
$= \left( \frac{\pi}{4} + \frac{1}{2} \right) - (0 + 1) = \frac{\pi}{4} - \frac{1}{2}$

5. Evaluate the definite integral for the second interval.

$\int_{-\frac{1}{2}}^{\frac{\sqrt{3}-2}{2}} \left( 1 - \frac{2}{4(x+1)^2+1} \right)\,dx$
$= \Big[ x - \arctan(2x+2) \Big]_{-\frac{1}{2}}^{\frac{\sqrt{3}-2}{2}}$
$= \left( \frac{\sqrt{3}-2}{2} - \arctan(\sqrt{3}) \right) - \left( -\frac{1}{2} - \arctan(1) \right)$
$= \left( \frac{\sqrt{3}}{2} - 1 - \frac{\pi}{3} \right) - \left( -\frac{1}{2} - \frac{\pi}{4} \right)$
$= \frac{\sqrt{3}}{2} - \frac{1}{2} - \frac{\pi}{12}$

6. Combine the results for the final exact answer.

$\text{Total Area} = \left( \frac{\pi}{4} - \frac{1}{2} \right) + \left( \frac{\sqrt{3}}{2} - \frac{1}{2} - \frac{\pi}{12} \right)$
$= \frac{\sqrt{3}}{2} - 1 + \frac{2\pi}{12} = \frac{\sqrt{3}}{2} - 1 + \frac{\pi}{6}$