Question

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:

• A. $(2,1)$
• B. $(-2,3)$
• C. $(2,3)$
• D. $(-2,1)$

Answer 1

The correct option is C $(2,3)$

$LetI=\int \frac{\tan x}{1+\tan x+{\tan }^{2}x}dxLet\tan x=u\frac{du}{dx}={\sec }^{2}x=1+u^{2}Now,I=\int \frac{u}{\left(1+u+u^2\right)\left(1+u^2\right)}du=\int \frac{\left(1+u+u^2\right)-\left(1+u^2\right)}{\left(1+u+u^2\right)\left(1+u^2\right)}du={\tan }^{-1}u-\int \frac{1}{1+u+u^2}du={\tan }^{-1}u-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{u+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C={\tan }^{-1}\left(\tan x\right)-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+C=x-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+C$

On comparing, we get

$k=2A=3$

$Hence,optionCisthecorrectanswer.$