Phân tích $\dfrac{1}{2-3x^2} =\dfrac{1}{(\sqrt 2 -\sqrt 3x )(\sqrt 2 +\sqrt 3 x)} =\dfrac{1}{2\sqrt 2}\bigg (\dfrac{1}{\sqrt 2 -\sqrt 3 x}+\dfrac{1}{\sqrt 2 +\sqrt 3 x} \bigg )$
$I = \dfrac{1}{2\sqrt 2}\int \dfrac{1}{\sqrt 2 -\sqrt 3 x}dx+ \dfrac{1}{2\sqrt 2}\int \dfrac{1}{\sqrt 2 +\sqrt 3 x}dx$
$= -\dfrac{\sqrt 3}{2\sqrt 2}\int \dfrac{d(\sqrt 2 -\sqrt 3 x)}{\sqrt 2 -\sqrt 3 x}+ \dfrac{\sqrt 3}{2\sqrt 2}\int \dfrac{d(\sqrt 2 +\sqrt 3 x)}{\sqrt 2 +\sqrt 3 x}$
$=\dfrac{\sqrt 3}{2\sqrt 2} \bigg (\ln |\sqrt 2 +\sqrt 3 x| - \ln |\sqrt 2 -\sqrt 3 x| \bigg )+C = \dfrac{\sqrt 3}{2\sqrt 2} \ln \bigg | \dfrac{\sqrt 2 +\sqrt 3 x}{\sqrt 2 -\sqrt 3 x} \bigg | +C$