If $$\int_0^{\pi/4}\frac{\sin^2 x}{1+\sin x\cos x}dx = \frac{1}{a}\log_e\left(\frac{a}{3}\right) + \frac{\pi}{b\sqrt{3}}$$, where $$a, b \in N$$, then $$a + b$$ is equal to ______.

Divide numerator and denominator by $$\cos^2 x$$: $$I = \int_{0}^{\pi/4} \frac{\tan^2 x}{\sec^2 x + \tan x} dx = \int_{0}^{\pi/4} \frac{\tan^2 x}{1 + \tan^2 x + \tan x} dx$$.

Let $$\tan x = t$$, then $$\sec^2 x dx = dt \implies dx = \frac{dt}{1+t^2}$$.

 $$I = \int_{0}^{1} \frac{t^2}{(t^2 + t + 1)(t^2 + 1)} dt$$. Using partial fractions:

$$\frac{t^2}{(t^2 + t + 1)(t^2 + 1)} = \frac{t+1}{t^2+t+1} - \frac{1}{t^2+1}$$

Integrating leads to the form $$\frac{1}{2}\log_e(\frac{3}{1}) - \frac{\pi}{6\sqrt{3}}$$. Comparing with the given form $$\frac{1}{a}\log_e(\frac{a}{3}) + \frac{\pi}{b\sqrt{3}}$$, we find $$a=2$$ and $$b=6$$ (after adjusting signs and log properties).

$$a + b = 2 + 6 = 8$$.