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$\int \frac{\cos x - 1}{\sin x + 1} e^{x} dx$ is equal to

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excos⁡x1+sin⁡x+C

C−exsin⁡x1+sin⁡x

C−ex1+sin⁡x

C−excos⁡x1+sin⁡x

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detailed solution

Correct option is A

∫cos⁡x−1sin⁡x+1exdx=∫cos⁡x1+sin⁡xexdx−∫11+sin⁡xexdx=(cos⁡x)ex1+sin⁡x−∫−(1+sin⁡x)sin⁡x−cos2⁡x(1+sin⁡x)2exdx−∫exsin⁡x+1dx+C =excos⁡x1+sin⁡x+∫11+sin⁡xexdx−∫ex1+sin⁡xdx+C=excos⁡x1+sin⁡x+C

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∫cos⁡x−1sin⁡x+1exdx is equal to

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$\int \frac{\cos x - 1}{\sin x + 1} e^{x} dx$ is equal to