Answer
$0$
Work Step by Step
Given: $I=\int_{-\sqrt 2}^{\sqrt 2} 3x \sqrt {2x^2+1} \ dx$
Let us consider that $u=2x^2+1 \implies dx=\dfrac{du}{4x}$
Now, we have $I=3 \int_{-\sqrt 2}^{\sqrt 2} x u^{1/2} (\dfrac{du}{4x})$
or, $=\dfrac{3}{4} \int_{-\sqrt 2}^{\sqrt 2} u^{1/2} \ du$
or, $=\dfrac{1}{2} [\dfrac{u^{3/2}}{3/2}]_{-\sqrt 2}^{\sqrt 2} +C$
or, $=\dfrac{1}{2}[(2x^2+1)^{3/2}]_{-\sqrt 2}^{\sqrt 2} $
or, $=\dfrac{1}{2}[(2(\sqrt 2)^2+1)^{3/2}]-\dfrac{1}{2}[(2(-\sqrt 2)^2+1)^{3/2}$
or, $=0$
