\(\int \frac{dx}{1+\tan x}\) equals 

\(\frac{x}{2}+\frac{1}{2}\log \left|\cos x+\sin x\right|+C\)

\(I = \int{\frac{dx}{1 + tanx}}  = \int{\frac{dx}{1 + \frac{sinx}{cosx}}} = \int{\frac{cosx}{sinx + cos x}dx} \)

\(= \frac{1}{2}\int{\frac{(cosx - sinx) + (cosx + sin x)}{cosx + sinx}dx}\)

\(=\frac{1}{2}\int{dx} + \frac{1}{2}\int{\frac{cosx − sinx}{cosx + sinx}dx}\)

Let \(cosx + sinx = t\)

\(⇒ dt = (cosx − sinx)dx\)

\(⇒I = \frac{x}{2} + \frac{1}{2}\int{\frac{dt}{t}} = \frac{x}{2} + \frac{log|t|}{2} + c\)

\(= \frac{x}{2} + \frac{1}{2}\begin{vmatrix}cosx + sinx\end{vmatrix} + c\)