$\int {{e^{2x}}( - \sin x + 2\cos x)\,dx} = $

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

$\int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x=$

If $\int e^x(\sin^2 2x - 8 \cos 4x) dx = e^x f(x) + c$, then $f(\frac{\pi}{4}) = $

$\int e^x \cdot \sec x(1+\tan x) \, dx = $ . . . . . . $+ C$.

Let $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$. Then find $\int e^x(f(x) + f'(x)) dx$ (where $c$ is the constant of integration).