Question

If $\int \frac{\tan x}{1+\tan x+{\tan }^{2}x}dx=x-\frac{2}{\sqrt{A}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{A}}\right)+C,$ where $C$ is arbitrary constant of integration, then the value of $A$ is

• A. 3.00
• B. 3.0
• C. 3
• D. 3.000

Answer 1

Answer: (A), (B), (C), (D)

$I=\int \frac{\tan x}{1+\tan x+{\tan }^{2}x}dx=\int \frac{\frac{\sin x}{\cos x}}{\frac{1}{{\cos }^{2}x}+\frac{\sin x}{\cos x}}dx=\int \frac{\sin 2x}{2+\sin 2x}dx=\int \frac{2+\sin 2x-2}{2+\sin 2x}dx=\int dx-2\int \frac{dx}{2+\sin 2x}$

Dividing numerator and denominator by ${\cos }^{2}x$ in second integral,
$I=x-2\int \frac{{\sec }^{2}x}{2{\sec }^{2}x+2\tan x}dx\Rightarrow I=x-I_1$
In integral $I_1,$ Put $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt\Rightarrow I_1=\frac{2}{2}\int \frac{dt}{t^2+t+1}\Rightarrow I_1=\int \frac{dt}{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\Rightarrow I_1=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t+1}{\sqrt{3}}\right)+C[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C]\Rightarrow I=x-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+C\therefore A=3$