$\quad \displaystyle\int\dfrac{1}{2x^2 +x +1}dx$
$= \displaystyle\int\dfrac{1}{\left(x\sqrt2+\dfrac{1}{2\sqrt2}\right)^2 +\dfrac78}dx$
Đặt $u = x\sqrt2+\dfrac{1}{2\sqrt2}$
$\to du = \sqrt2 dx$
Ta được:
$\quad \dfrac{1}{\sqrt2}\displaystyle\int\dfrac{1}{u^2 +\dfrac78}du$
$= \dfrac{4\sqrt2}{7}\displaystyle\int\dfrac{1}{\dfrac{8u^2}{7} + 1}du$
Đặt $t = 2u\sqrt{\dfrac27}$
$\to dt = 2\sqrt{\dfrac27}du$
Ta được:
$= \dfrac{2}{\sqrt7}\displaystyle\int\dfrac{1}{t^2 +1}dt$
$= \dfrac{2}{\sqrt7}\cdot \arctan t + C$
$= \dfrac{2}{\sqrt7}\cdot \arctan\left(2u\sqrt{\dfrac27}\right)+ C$
$= \dfrac{2}{\sqrt7}\cdot \arctan\left(\dfrac{4x+1}{\sqrt7}\right)+ C$
