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

Answer: (C)

Given:
$I=\int \frac{\tan }{1+\tan x+{\tan }^{2}x}dx=\int \left(1-\frac{{\sec }^{2}x}{1+\tan x+{\tan }^{2}x}\right)dx$
Let $\tan x=t,{\sec }^{2}xdx=dt$
So, $I=x-\int \frac{dt}{1+t+t^2}$
$=x-\int \frac{dt}{{\left(t+\frac{1}{2}\right)}^2+\frac{3}{4}}$
$=x-\frac{1}{\sqrt{3}/2}{\tan }^{-1}\left(\frac{t+\frac{1}{2}}{\sqrt{3}/2}\right)+C$
$=x-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+C$
Therefore, values of $K=2,A=3$