Đáp án:
$D.\, 12$
Giải thích các bước giải:
Xét $I = \displaystyle\int\dfrac{1}{x^2 + x +1}dx$
$\begin{array}{l} \to I = \displaystyle\int\dfrac{1}{\left(x + \dfrac12\right)^2 + \dfrac{3}{4}}dx\\ Đặt\,\,u = x + \dfrac12\\ \to du = dx\\ \text{Ta được:}\\ \quad I = \displaystyle\int\dfrac{1}{u^2 + \dfrac34}du\\ \to I = \dfrac{4}{3}\displaystyle\int\dfrac{1}{\dfrac{4}{3}u^2 + 1}du\\ Đặt\,\,t = \dfrac{2}{\sqrt3}u\\ \to dt = \dfrac{2}{\sqrt3}du\\ \text{Ta được:}\\ \quad I = \dfrac{2}{\sqrt3}\displaystyle\int\dfrac{1}{t^2 + 1}dt\\ \to I = \dfrac{2}{\sqrt3}\arctan t + C\\ \to I = \dfrac{2}{\sqrt3}\arctan\left(\dfrac{2}{\sqrt3}u\right)+C\\ \to I = \dfrac{2}{\sqrt3}\arctan\left(\dfrac{2x+1}{\sqrt3}\right)+C \end{array}$
Khi đó:
$\quad \displaystyle\int\limits_0^1\dfrac{1}{x^2 + x +1}dx$
$= \dfrac{2}{\sqrt3}\arctan\left(\dfrac{2x+1}{\sqrt3}+ C\right)\Bigg|_0^1$
$= \dfrac{2}{\sqrt3}\left(\arctan\sqrt3- \arctan\dfrac{\sqrt3}{3}\right)$
$=\dfrac{2}{\sqrt3}\left(\dfrac{\pi}{3} -\dfrac{\pi}{6}\right)$
$=\dfrac{\pi\sqrt3}{9}$
$\to \begin{cases}a = 3\\b = 9\end{cases}$
$\to a + b = 12$
