$\int \frac{d x}{1+\tan x}$ is equal to:

$\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+c$, C is an arbitrary constant

The correct answer is Option (2) → $\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+c$, C is an arbitrary constant

$\int \frac{d x}{1+\tan x}=\int\frac{\cos x}{\cos x+\sin x}dx$

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

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

$=\frac{x}{2}+\frac{1}{2}\log|\cos x+\sin x|+c$