Đáp án:
$B.\, \log_2\dfrac{\alpha}{\beta}=2$
Giải thích các bước giải:
$\begin{array}{l}+) \quad \displaystyle\int\limits_{-1}^0\dfrac{1}{x^2 + 2x + 2}dx\\ = \displaystyle\int\limits_{-1}^0\dfrac{1}{(x+1)^2 + 1}d(x+1)\\ = \arctan(x+1)\Bigg|_{-1}^0\\ = \dfrac{\pi}{4}\\ \to \alpha = \dfrac{\pi}{4}\\ +)\quad \displaystyle\int\limits_0^1\dfrac{x^3}{x^8 + 1}dx\\ = \dfrac{1}{4}\displaystyle\int\limits_0^1\dfrac{1}{(x^4)^2 + 1}d(x^4)\\ = \dfrac{1}{4}\arctan(x^4)\Bigg|_0^1\\ =\dfrac{\pi}{16}\\ \to \beta = \dfrac{\pi}{16}\\ \to \dfrac{\alpha}{\beta} =\dfrac{\dfrac{\pi}{4}}{\dfrac{\pi}{16}} = 4\\ \to \log_2\dfrac{\alpha}{\beta}=\log_24=2 \end{array}$
