$\int_1^{e} \frac{e^x}{x}(1+x \log x) d x=$

The value of the integral $\int_{0}^{\infty} e^{-2x} (\sin 2x + \cos 2x) dx$ is:

If $I = \int e^{\sin \theta} (\log \sin \theta + \operatorname{cosec}^2 \theta) \cos \theta \, d\theta$,then $I$ is equal to

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

Let $f(t) = \int \left( \frac{1 - \sin(\ln t)}{1 - \cos(\ln t)} \right) dt$,for $t > 1$. If $f(e^{\pi/2}) = -e^{\pi/2}$ and $f(e^{\pi/4}) = \alpha e^{\pi/4}$,then $\alpha$ equals:

$\int {e^x \frac{x^2 + 1}{(x + 1)^2} dx} = $