Answer
$$\displaystyle{\int{\frac{1+\sin x}{1+\cos x}dx}=\csc x-\cot x-\ln \left| 1+\cos x \right|+C}$$
Work Step by Step
$\displaystyle{I=\int{\frac{1+\sin x}{1+\cos x}dx}\\
I=\int{\frac{1+\sin x}{1+\cos x}\times \frac{1-\cos x}{1-\cos x}dx}\\
I=\int{\frac{1-\cos x+\sin x-\cos x\sin x}{1-\cos ^2x}dx}\\
I=\int{\frac{1-\cos x+\sin x-\cos x\sin x}{\sin ^2x}dx}\\
I=\int{\frac{1}{\sin ^2x}-\frac{\cos x}{\sin ^2x}+\frac{\sin x}{\sin ^2x}-\frac{\cos x\sin x}{\sin ^2x}dx}\\
I=\int{\csc^2x-\cot x\csc x+\csc x-\cot x}\ dx\\
I=-\cot x+\csc x-\ln \left| \csc x+\cot x \right|-\ln \left| \sin x \right|+C\\
I=\csc x-\cot x-\ln \left| 1+\cos x \right|+C}
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