Answer
$\dfrac{5 \sqrt 5-1}{6}$
Work Step by Step
Given: $I=\int_{0}^{\sqrt 2} x \sqrt {2x^2+1} \ dx$
Let us consider that $u=2x^2+1 \implies dx=\dfrac{du}{4x}$
Now, we have $I=\int_{0}^{\sqrt 2} x u^{1/2} (\dfrac{du}{4x})$
or, $=\dfrac{1}{4} \int_{0}^{\sqrt 2} u^{1/2} \ du$
or, $=\dfrac{1}{4} [\dfrac{u^{3/2}}{3/2}]_{0}^{\sqrt 2} +C$
or, $=\dfrac{1}{6}[(2x^2+1)^{3/2}]_{0}^{\sqrt 2} $
or, $=\dfrac{1}{6}[(2(\sqrt 2)^2+1)^{3/2}]-\dfrac{1}{6}[(2(0)^2+1)^{3/2}]$
or, $=\dfrac{5 \sqrt 5-1}{6}$
