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

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$e^{x} + \sin x + 2 x$

$e^{x} + \sin x$

$e^{x} - \sin x$

$e^{x} + \sin x + x$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int \frac{(e^{x} + \cos x + 1) - (e^{x} + \sin x + x)}{e^{x} + \sin x + x} dx \\ = \log_{e} (e^{x} + \sin x + x) - x + C \\ ∴ f (x) = e^{x} + \sin x + x and g (x) = - x \\ f (x) + g (x) = e^{x} + \sin x \end{array}$

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