If an antiderivative of $f(x)$ is $e^x$ and an antiderivative of $g(x)$ is $\cos x$, then $\int f(x)\cos x\, dx + \int g(x)e^x\, dx = $

$\int_0^{\log 10}\left[\dfrac{e^x \sqrt{e^x - 1}}{e^x + 8}\right]dx = $

The value of the integral $\int_{-\pi/2}^{\pi/2}\left(\dfrac{x^2 \cos x}{1+e^x}\right)dx$ is equal to $\left(\dfrac{\pi^2}{A}\right) - B$. Then $\left(\dfrac{A}{B}\right) = $

The area of the smaller region bounded by the circle $x^2 + y^2 = 4$ and the line $x = 1$ is...