$\int_{\frac{1}{25}}^3 \frac{e^{\frac{3}{x}}}{x^2} d x=$

$\int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{\frac{1}{3}}} dx$ is equal to

$\int_0^{\pi /6} \frac{\sin x}{\cos^3 x} \, dx = $

$\int_{-3}^0 x \sqrt{x+4} \, dx =$

Let $y=f(x)$ be a thrice differentiable function in $(-5,5)$. Let the tangents to the curve $y=f(x)$ at $(1, f(1))$ and $(3, f(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$,respectively with the positive $x$-axis. If $27 \int_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3}$,where $\alpha$ and $\beta$ are integers,then the value of $\alpha+\beta$ equals:

$\int\limits_0^{{{\left( {\frac{\pi }{2}} \right)}^{\frac{1}{3}}}} {\,{x^5}\cdot\sin {x^3}\,dx} $ $=$