If int {frac{{tan x}}{{1 + tan x + {{tan }^2} | vedclass

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(A) Let $I = \int \frac{	an x}{1 + 	an x + 	an^2 x} dx$.We can rewrite the numerator as $\frac{1}{2} [ (1 + 	an x + 	an^2 x) - (1 - 	an x + 	an

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$(2, 3)$

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(A) Let $I = \int \frac{	an x}{1 + 	an x + 	an^2 x} dx$.We can rewrite the numerator as $\frac{1}{2} [ (1 + 	an x + 	an^2 x) - (1 - 	an x + 	an^2 x) ]$ is not ideal,so let's use the identity: $	an x = \frac{1}{2} (1 + 	an x + 	an^2 x) - \frac{1}{2} (1 - 	an x + 	an^2 x)$.Alternatively,note that $\int \frac{	an x}{1 + 	an x + 	an^2 x} dx = \int \frac{	an x + 1 + 	an^2 x - (1 + 	an^2 x)}{1 + 	an x + 	an^2 x} dx = \int 1 dx - \int \frac{\sec^2 x}{1 + 	an x + 	an^2 x} dx$.Let $I = x - \int \frac{\sec^2 x}{1 + 	an x + 	an^2 x} dx$.Substitute $	an x = t$,so $\sec^2 x dx = dt$.$I = x - \int \frac{dt}{t^2 + t + 1} = x - \int \frac{dt}{(t + 1/2)^2 + 3/4}$.Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = x - \frac{1}{\sqrt{3}/2} 	an^{-1} \left( \frac{t + 1/2}{\sqrt{3}/2} ight) + C = x - \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2t + 1}{\sqrt{3}} ight) + C$.Substituting $t = 	an x$,we have $I = x - \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2 	an x + 1}{\sqrt{3}} ight) + C$.Comparing this with the given form $x - \frac{K}{\sqrt{A}} 	an^{-1} \left( \frac{K 	an x + 1}{\sqrt{A}} ight) + C$,we find $K = 2$ and $A = 3$.Thus,the ordered pair $(K, A)$ is $(2, 3)$.

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

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