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(A) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.Given integral is $I = \int e^x \left( \frac{1+\sin x}{1+\cos x} \right) dx$.Using trigon

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$e^x 	an \frac{x}{2} + c$

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(A) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.Given integral is $I = \int e^x \left( \frac{1+\sin x}{1+\cos x} ight) dx$.Using trigonometric identities $1+\cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:$I = \int e^x \left( \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} ight) dx$$I = \int e^x \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} ight) dx$$I = \int e^x \left( \frac{1}{2} \sec^2 \frac{x}{2} + 	an \frac{x}{2} ight) dx$Let $f(x) = 	an \frac{x}{2}$. Then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$.Since the integral is in the form $\int e^x [f(x) + f'(x)] dx$,the result is $e^x f(x) + c$.Therefore,$I = e^x 	an \frac{x}{2} + c$.

$\int e^x \left( \frac{1+\sin x}{1+\cos x} \right) dx =$

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