Answer
$$\tan x-\sec x+c$$
Work Step by Step
Given$$\int \frac{dx}{1+\sin x}$$
Since
\begin{align*}
\int \frac{dx}{1+\sin x}\frac{1-\sin x}{1-\sin x}
&=\int \frac{(1-\sin x)dx}{1-\sin^2 x}\\
&=\int \frac{(1-\sin x)dx}{\cos^2 x}\\
&=\int [\sec^2 x-\sec x\tan x]dx\\
&=\tan x-\sec x+c
\end{align*}
