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

Changing the limits: when $$x = 0 \implies t = 0$$, and when $$x = \frac{\pi}{4} \implies t = 1$$:

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

Using partial fraction decomposition:

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

Split the integral into two parts ($$I = I_1 - I_2$$):

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

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

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

Express the numerator as $$t = \frac{1}{2}(2t + 1) - \frac{1}{2}$$:

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

$$\frac{1}{2} \left[ \log_e(t^2 + t + 1) \right]_0^1 = \frac{1}{2} (\log_e 3 - \log_e 1) = \frac{1}{2} \log_e 3$$

$$t^2 + t + 1 $$= $$\left(t + \frac{1}{2}\right)^2 + \frac{3}{4} = \left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$

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

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

$$I = I_1 - I_2$$ 

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

Comparing with the given form $$\frac{1}{a}\log_e\left(\frac{a}{3}\right) + \frac{\pi}{b\sqrt{3}}$$, we find:

$$a = 2 \quad \text{and} \quad b = 6$$ $$a + b = 2 + 6 = 8$$