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

Answer 1

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

$\int \frac{\tan x+1+{\tan }^{2}x}{\tan x+1+{\tan }^{2}x}dx-\int \frac{(1+{\tan }^{2}x)}{1+\tan +{\tan }^{2}x}dx$

$=x-\int \frac{{\sec }^2+dx}{1+\tan x+{\tan }^{2}x}$

$=x-\int \frac{dt}{t^2+t+\frac{1}{4}+1-\frac{1}{4}}$

$=x-\int \frac{dt}{{(t+\frac{1}{2})}^2+{\left(\frac{3}{2}\right)}^2}$

$=x-\frac{2}{c_3}{\tan }^{-}\left(\frac{t+{11}_2}{c_3/2}\right)+c$

$x-\frac{2}{c_3}{\tan }^{-}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+c$

$A=3$

$K=2$.

Thus the answer is $(2,3)$