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

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

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

$$I = \int_{0}^{1} \frac{t}{t^2 + 1} \, dt - \int_{0}^{1} \frac{t}{t^2 + t + 1} \, dt$$

$$I_1 = \int_{0}^{1} \frac{t}{t^2 + 1} \, dt = \frac{1}{2} \left[ \log_e(t^2 + 1) \right]_0^1 = \frac{1}{2} (\log_e 2 - \log_e 1) = \frac{1}{2} \log_e 2$$

$$I_2 = \int_{0}^{1} \frac{t}{t^2 + t + 1} \, dt$$

$$t = \frac{1}{2}(2t + 1) - \frac{1}{2}$$ $$I_2 = \frac{1}{2} \int_{0}^{1} \frac{2t + 1}{t^2 + t + 1} \ dt - \frac{1}{2} \int_{0}^{1} \frac{1}{t^2 + t + 1} \ dt$$
$$\frac{1}{2} \left[ \log_e(t^2 + t + 1) \right]_0^1 = \frac{1}{2} (\log_e 3 - \log_e 1) = \frac{1}{2} \log_e 3$$
$$t^2 + t + 1 = \left(t + \frac{1}{2}\right)^2 + \frac{3}{4} = \left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$

$$\frac{1}{2} \int_{0}^{1} \frac{1}{\left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dt = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) \right]_0^1$$ $$= \frac{1}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{2t + 1}{\sqrt{3}}\right) \right]_0^1 = \frac{1}{\sqrt{3}} \left( \tan^{-1}(\sqrt{3}) - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \right)$$ $$= \frac{1}{\sqrt{3}} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{6\sqrt{3}}$$

$$I_2 = \frac{1}{2} \log_e 3 - \frac{\pi}{6\sqrt{3}}$$

$$I = I_1 - I_2$$$$I = \frac{1}{2} \log_e 2 - \left( \frac{1}{2} \log_e 3 - \frac{\pi}{6\sqrt{3}} \right)$$ 

$$I = \frac{1}{2} (\log_e 2 - \log_e 3) + \frac{\pi}{6\sqrt{3}}$$$$I = \frac{1}{2} \log_e\left(\frac{2}{3}\right) + \frac{\pi}{6\sqrt{3}}$$

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

Comparing with the given form $$\frac{1}{a}\log_e(\frac{a}{3}) + \frac{\pi}{b\sqrt{3}}$$, we find $$a=2$$ and $$b=6$$ 

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