Evaluate the definite integral $\int_{0}^{1} \frac{d x}{1+x^{2}}$.

$\int_{-1}^4 \sqrt{\frac{4-x}{x+1}} \, dx =$

The value of the integral $\int_0^4 \frac{d x}{1+x^2}$ obtained by using the Trapezoidal rule with $h=1$ is

If $\int_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{(1+x^{2})^{3}}}} dx = \alpha \sqrt{2} + \beta \sqrt{3}$,where $\alpha, \beta$ are integers,then $\alpha + \beta$ is equal to.

If $g(x) = \int_{0}^{x} \cos 4t \, dt$,then $g(x + \pi) = $

$\int_{-2}^2 |[x]| \, dx$ is equal to