Ta có:
$\int\dfrac{1}{x^2+a}dx=\dfrac{1}{\sqrt a}\arctan(\dfrac{x}{\sqrt a})+C$ (tự chứng minh).
Ta có:
$\int\limits_0^1\dfrac{1}{x^4+4x^2+3}dx$
$=\dfrac{1}{2}\int\limits_0^1\left(\dfrac{1}{x^2+1}-\dfrac{1}{x^2+3}\right)dx$
$=\left(\dfrac{1}{2}\arctan x-\dfrac{1}{2\sqrt3}\arctan(\dfrac{x}{\sqrt 3})\right)\left|\begin{array}{l}1\\0\end{array}\right.$
$=\dfrac{(9-2\sqrt3)\pi}{72}$