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

$\int e^x \left(\frac{x+2}{x+4}\right)^2 dx =$

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).

If $f(x) = \frac{1}{\log x}$ and $g(x) = \frac{1}{(\log x)^2}$, then the value of $\int [f(x) - g(x)] dx$ is...

$\int \frac{(x-1) e^{x}}{(x+1)^{3}} d x$ is equal to

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: