$\int\limits^1_0 \frac{1}{x^2+2} \, dx$

Đáp án:

Giải thích các bước giải:

$x=\sqrt{2}tan(t)\\ t=\arctan\left(\dfrac{x}{\sqrt{2}}\right)\\ x \quad | \quad 0 \quad | \quad 1 \quad\\ \overline{t \quad | \quad 0 \quad | \quad \arctan\left(\dfrac{1}{\sqrt{2}}\right) \quad}\\ dx=\sqrt{2}(tan(t)^2+1)dt\\ \displaystyle\int\limits_0^{\arctan\left(\dfrac{1}{\sqrt{2}}\right)}\dfrac{\sqrt{2}(tan(t)^2+1)dt}{2(tan(t)^2+1)}\\ =\displaystyle\int\limits_0^{\arctan\left(\dfrac{1}{\sqrt{2}}\right)}\dfrac{1}{\sqrt{2}}dt\\ =\dfrac{1}{\sqrt{2}}t\Bigg\vert_0^{\arctan\left(\dfrac{1}{\sqrt{2}}\right)}\\ =\dfrac{1}{\sqrt{2}}\arctan\left(\dfrac{1}{\sqrt{2}}\right)$