Question

If $\int \frac{\tan x}{1+\tan x+\tan ^{2} x} d x=x-\frac{K}{\sqrt{A} \tan ^{-1}\left(\frac{K \tan x+1}{\sqrt{A}}\right)}+C$ (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:
$I=\int \frac{\tan }{1+\tan x+\tan ^{2} x} d x=\int\left(1-\frac{\sec ^{2} x}{1+\tan x+\tan ^{2} x}\right) d x$
Let $\tan x=t, \sec ^{2} x d x=d t $
So, $ I=x-\int \frac{d t}{1+t+t^{2}}$
$=x-\int \frac{d t}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}}$
$=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 $K=2, A=3$