Answer
$$\frac{\left(4a^2+1\right)\left(2a+1\right)}{4a^2\left(2a-1\right)}$$
Work Step by Step
Recall, one never divides by a fraction. Rather, if we want to divide by a fraction, we multiply by the reciprocal of the fraction. Recall, the reciprocal of a fraction is that fraction with the numerator and the denominator switched. Thus, we find:
$$ \frac{4a^2+1}{4a^2-1}\times \frac{4a^2+4a+1}{4a^2}\\ \frac{\left(4a^2+1\right)\left(2a+1\right)^2}{\left(2a+1\right)\left(2a-1\right)\times \:4a^2}\\ \frac{\left(4a^2+1\right)\left(2a+1\right)}{4a^2\left(2a-1\right)}$$
