If $\int \frac{cos x - sin x + 1 - x}{e^{x} + sin x + x} d x - ln (f (x)) + g (x) + C$ where $C$ is the constant of integration and $f (x)$ is positive , then $f (x) + g (x)$ has the value equal to

e^{x} + sin x + 2 x

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e^{x} + cos x + 1

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e^{x} - sin x

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e^{x} + sin x + x

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Solution

The correct option is A $e^{x} + cos x + 1$
$I = \int \frac{cos x - sin x + 1 - x}{e^{x} + sin x + x} d x$
$= \int \frac{e^{x} + cos x + 1 - e^{x} - sin x - x}{e^{x} + sin x + x} d x$
$= \int (\frac{e^{x} + cos x + 1}{e^{x} + sin x + x} - 1) d x$
$= log (e^{2} + sin x + x) - x$
$∵ \frac{d}{d x} (e^{x} + sin x + x) = e^{x} + cos x + 1$

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