Question

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

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

Answer 1

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

$I=\int \frac{\tan x}{1+\tan x+{\tan }^{2}x}dx=\int (1-\frac{{\sec }^{2}x}{1+\tan x+{\tan }^{2}x})dx=x-\int \frac{1}{1+t+t^2}dt(\text{put}\tan x=t)=x-\int \frac{1}{(t+\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}dt=x-\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c=x-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+c\Rightarrow K=2\&A=3$