Đáp án:
$2\sqrt{13}-2.$
Giải thích các bước giải:
$\displaystyle\int\limits^3_0 \dfrac{2x+1}{\sqrt{x^2+x+1}} \, dx\\ =\displaystyle\int\limits^3_0 \dfrac{d(x^2+x)}{\sqrt{x^2+x+1}} \, dx\\ =\displaystyle\int\limits^3_0 \dfrac{d(x^2+x+1)}{\sqrt{x^2+x+1}} \, dx\\ =2\displaystyle\int\limits^3_0 \dfrac{d(x^2+x+1)}{2\sqrt{x^2+x+1}} \, dx\\ =2\sqrt{x^2+x+1} \Bigg\vert^3_0\\ =2\sqrt{3^2+3+1}-2\sqrt{0^2+0+1}\\ =2\sqrt{13}-2.$
