If ∫tan⁡x1+tan⁡x+tan2⁡xdx=x−KAtan−1⁡Ktan⁡x+1A+C,(C is a constant of integration), then the ordered pair (K,A) is equal to

Let I=∫tan⁡x1+tan⁡x+tan2⁡xdx⇒I=∫tan⁡x+1+tan2⁡xtan⁡x+1+tan2⁡xdx−∫1+tan2⁡x1+tan⁡x+tan2⁡xdx⇒I=x−∫sec2⁡x1+tan⁡x+tan2⁡xdx Put tan⁡x=t⇒sec2⁡x⋅dx=dt∴I=x−∫dtt2+t+1∴I=x−∫dtt2+t+14+1−14=x−∫dtt+122+322⇒I=x−23tan−1⁡t+1232+C⇒I=x−23tan−1⁡2tan⁡x+13+C∴A=3 and K=2

If ∫tan⁡x1+tan⁡x+tan2⁡xdx=x−KAtan−1⁡Ktan⁡x+1A+C,(C is a constant of integration), then the ordered pair (K,A) is equal to

Let I=∫tan⁡x1+tan⁡x+tan2⁡xdx⇒I=∫tan⁡x+1+tan2⁡xtan⁡x+1+tan2⁡xdx−∫1+tan2⁡x1+tan⁡x+tan2⁡xdx⇒I=x−∫sec2⁡x1+tan⁡x+tan2⁡xdx Put tan⁡x=t⇒sec2⁡x⋅dx=dt∴I=x−∫dtt2+t+1∴I=x−∫dtt2+t+14+1−14=x−∫dtt+122+322⇒I=x−23tan−1⁡t+1232+C⇒I=x−23tan−1⁡2tan⁡x+13+C∴A=3 and K=2

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

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$(2, 3)$

$(2, 1)$

$(- 2, 1)$

$(- 2, 3)$

answer is A.

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Detailed Solution

$\begin{array}{l} Let I = \int \frac{\tan x}{1 + \tan x + \tan^{2} x} d x \\ \Rightarrow I = \int \frac{\tan x + 1 + \tan^{2} x}{\tan x + 1 + \tan^{2} x} d x - \int \frac{(1 + \tan^{2} x)}{1 + \tan x + \tan^{2} x} d x \\ \Rightarrow I = x - \int \frac{\sec^{2} x}{1 + \tan x + \tan^{2} x} d x \\ Put \tan x = t \\ \Rightarrow \sec^{2} x \cdot d x = d t \\ ∴ I = x - \int \frac{d t}{t^{2} + t + 1} \end{array}$

$\begin{array}{l} ∴ I = x - \int \frac{d t}{t^{2} + t + \frac{1}{4} + 1 - \frac{1}{4}} \\ = x - \int \frac{d t}{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} \\ \Rightarrow I = x - \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C \\ \Rightarrow I = x - \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan x + 1}{\sqrt{3}}) + C \\ ∴ A = 3 and K = 2 \end{array}$