If (\int \frac{\tan x}{1 + \tan x + \tan 2x} dx = x - K\sqrt{A} \tan^{-1}(K \tan x + \frac{1}{\sqrt{A}}) + C), (C is a constant of integration), then the ordered pair (K, A) is equal to.

The correct option is A (2,3)
(\int \frac{\tan x}{1 + \tan x + \tan 2x} dx
\int \frac{\tan x}{1 + \tan x + \frac{2 \tan x}{1 - \tan^2 x}} dx = \int \frac{\tan x (1 - \tan^2 x)}{1 + \tan x + 2 \tan x - \tan^2 x - \tan^3 x} dx = \int \frac{\tan x (1 - \tan^2 x)}{1 + 3 \tan x - \tan^2 x - \tan^3 x} dx
\int \frac{\tan x}{1 + \tan x + \tan 2x} dx = x - \int \frac{1 + \tan^2 x}{1 + \tan x + \tan 2x} dx = x - \int \frac{sec^2 x}{1 + \tan x + \tan 2x} dx
Let ( t = \tan x ), then ( dt = \sec^2 x dx
\int \frac{\tan x}{1 + \tan x + \tan 2x} dx = x - \int \frac{dt}{1 + t + 2t/(1-t^2)} = x - \int \frac{dt}{1 + t + 2t/(1 - t^2)} = x - \int \frac{(1-t^2) dt}{1 + 3t - t^2 - t^3} = x - \int \frac{dt}{t^2 + t + 1} = x - \int \frac{dt}{(t + 1/2)^2 + (\sqrt{3}/2)^2}
= x - \frac{2}{\sqrt{3}} \tan^{-1}(\frac{2t + 1}{\sqrt{3}}) + C = x - \frac{2}{\sqrt{3}} \tan^{-1}(\frac{2 \tan x + 1}{\sqrt{3}}) + C
Comparing with ( x - K \sqrt{A} \tan^{-1}(K \tan x + \frac{1}{\sqrt{A}}) + C ), we have
K = 2, A = 3
Thus the answer is (2,3)