Ta có
$\displaystyle\int\limits_0^1 \dfrac{x}{4-x^2} dx = \displaystyle\int\limits_0^1 \dfrac{x}{(2-x)(2+x)}dx$
$=\dfrac{1}{2} \displaystyle\int\limits_0^1\left( \dfrac{1}{2-x} - \dfrac{1}{2+x} \right) dx$
$= \dfrac{1}{2} (\ln |2-x| - \ln|2+x|) \Bigg\vert_0^1$
$= \dfrac{1}{2} (\ln 1 - \ln 3) - \dfrac{1}{2} (\ln 2 - \ln 2)$
$= -\dfrac{1}{2} \ln 3$
Vậy
$\displaystyle\int\limits_0^1 \dfrac{x}{4-x^2} dx =-\dfrac{1}{2} \ln 3$
