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

We have to evaluate the integral $$\displaystyle\int \frac{\tan x}{1+\tan x+\tan^2x}\,dx$$ and then identify the constants $$K$$ and $$A$$ in the given result

First we make the standard substitution $$t=\tan x$$. Differentiating, we obtain $$dt=\sec^2x\,dx=\left(1+\tan^2x\right)\,dx=\left(1+t^2\right)\,dx$$, so that $$dx=\dfrac{dt}{1+t^2}$$.

Substituting these in the integral, we get

$$\int \frac{\tan x}{1+\tan x+\tan^2x}\,dx =\int \frac{t}{1+t+t^2}\;\frac{dt}{1+t^2} =\int\frac{t}{(1+t+t^2)(1+t^2)}\,dt.$$

According to the statement of the problem, the antiderivative is

$$x-\frac{K}{\sqrt{A}}\;\tan^{-1}\!\left(\frac{K\tan x+1}{\sqrt{A}}\right)+C.$$ Replacing $$x$$ by $$\tan^{-1}t$$ (because $$t=\tan x$$), the same result can be rewritten in terms of $$t$$ as

$$\tan^{-1}t-\frac{K}{\sqrt{A}}\;\tan^{-1}\!\left(\frac{K t+1}{\sqrt{A}}\right)+C.$$

In order for this expression to be an antiderivative, its derivative with respect to $$t$$ must equal the integrand we already found, namely $$\dfrac{t}{(1+t+t^2)(1+t^2)}$$.

We now differentiate the supposed antiderivative with respect to $$t$$. The derivative of $$\tan^{-1}u$$ is $$\dfrac{u'}{1+u^2}$$, so:

$$\frac{d}{dt}\Bigl[\tan^{-1}t\Bigr]=\frac{1}{1+t^2},$$

$$\frac{d}{dt}\Bigl[\tan^{-1}\!\bigl(\tfrac{K t+1}{\sqrt{A}}\bigr)\Bigr] =\frac{\,K/\sqrt{A}\,}{1+\bigl(K t+1\bigr)^2/A} =\frac{K\sqrt{A}}{A+\bigl(K t+1\bigr)^2}.$$

Putting these together, the derivative of the whole expression is

$$\frac{1}{1+t^2}-\frac{K}{\sqrt{A}}\cdot\frac{K\sqrt{A}}{A+(K t+1)^2} =\frac{1}{1+t^2}-\frac{K^2}{A+(K t+1)^2}.$$

We now bring both terms to a single fraction. The common denominator is $$\bigl(1+t^2\bigr)\bigl[A+(K t+1)^2\bigr]$$, so

$$\frac{1}{1+t^2}-\frac{K^2}{A+(K t+1)^2} =\frac{A+(K t+1)^2-K^2(1+t^2)}{\bigl(1+t^2\bigr)\bigl[A+(K t+1)^2\bigr]}.$$

Expanding the numerator:

$$A+(K t+1)^2-K^2(1+t^2) =A+K^2t^2+2K t+1-K^2-K^2t^2 =(A+1-K^2)+2K t.$$

Thus the derivative becomes

$$\frac{(A+1-K^2)+2K t}{(1+t^2)\bigl[A+K^2t^2+2K t+1\bigr]}.$$

But $$A+K^2t^2+2K t+1=(1+t^2)(1+t+t^2)$$ is not automatically true, so we equate the derivative we just found with the required integrand and match coefficients. We therefore impose

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

Cross-multiplying gives

$$(A+1-K^2+2K t)\,(1+t+t^2)=t\,(A+K^2t^2+2K t+1).$$

Expanding the left side completely:

$$A+1-K^2+2K t$$

times

$$1+t+t^2$$

yields

$$\bigl(A+1-K^2\bigr)+\bigl(A+1-K^2+2K\bigr)t+\bigl(A+1-K^2+2K\bigr)t^2+2K t^3.$$

The right side expands to

$$\bigl(A+1\bigr)t+2K t^2+K^2 t^3.$$

Matching coefficients of like powers of $$t$$ we get four equations:

Constant term: $$A+1-K^2=0.$$

Coefficient of $$t$$: $$A+1-K^2+2K=A+1.$$

Coefficient of $$t^2$$: $$A+1-K^2+2K=2K.$$

Coefficient of $$t^3$$: $$2K=K^2.$$

From the constant term we have $$K^2=A+1.$$ From the $$t^3$$ term we get $$K^2=2K,$$ leading to $$K(K-2)=0.$$ Since $$K=0$$ would give a negative $$A$$ (namely $$A=-1$$) and a meaningless square root, we discard it and keep $$K=2.$$ Substituting $$K=2$$ into $$K^2=A+1$$ gives $$4=A+1,$$ whence $$A=3.$$

Thus the ordered pair is $$(K,A)=(2,3).$$

Option B lists exactly this pair. Hence, the correct answer is Option B.