The solution of $\frac{dy}{dx} = e^x(\sin x + \cos x)$ is

$\int e^{x / 2}\left(\frac{2+\sin x}{1+\cos x}\right) d x=$

The value of $\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x$ is equal to

$\int e^x \frac{x^2+1}{(x+1)^2} d x$ is equal to

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

If $\int {{e^{{x^2}}}\left( {2 - \frac{1}{{{x^2}}}} \right)dx = {e^{{x^2}}}f(x) + C} $ and $f\left( {\frac{1}{2}} \right) = 2$,then $f(1)$ is equal to (where $C$ is an arbitrary constant).