If $I = \int e^{x \sin x + \cos x} (\frac{x^{4} \cos^{3} x - x \sin x + \cos x}{x^{2} \cos^{2} x}) d x$ then I equals
(C is the constant of integration)

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$e^{x \sin x + \cos x} (x - \frac{\sec x}{x}) + C$

$e^{x \sin x + \cos x} (x \sin x - \frac{\cos x}{x}) + C$

$e^{x \sin x + \cos x} (\frac{x}{\tan x} - \frac{\sec x}{x}) + C$

$x e^{x \sin x + \cos x} - \int e^{x \sin x + \cos x} (1 - \frac{\cos x - x \sin x}{x^{2} \cos^{2} x}) d x$

answer is A, D.

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Detailed Solution

$I = \int e^{x \sin x + \cos x} {x^{2} \cos x + \frac{\cos x - x \sin x}{{(x \cos x)}^{2}}} d x$
$= I_{1} + I_{2}$
Where
   $I_{1} = \int (e^{x \sin x + \cos x} x \cos x) x d x$
Noting $\frac{d}{d x} (x \sin x + \cos x) = x \cos x,$ we get
And    $I_{1} = x e^{x \sin x + \cos x} - \int e^{x \sin x + \cos x} d x$
   $I_{2} = \int e^{x \sin x + \cos x} \frac{(\cos x - x \sin x)}{{(x \cos x)}^{2}} d x$
= $e^{x \sin x + \cos x} (- \frac{1}{x \cos x})$
+ $\int e^{x \sin x + \cos x} (x \cos x) \frac{d x}{x \cos x}$
Thus, $I = e^{x \sin x + \cos x} (x - \frac{1}{x \cos x}) + C$