Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\frac{x}{{\sqrt {1 + 2{x^2}} }}} dx \cr
& {\text{Integrating }}\int {\frac{x}{{\sqrt {1 + 2{x^2}} }}} dx \cr
& u = 1 + 2{x^2},{\text{ }}du = 4xdx,{\text{ }}dx = \frac{1}{{4x}}du \cr
& \int {\frac{x}{{\sqrt {1 + 2{x^2}} }}} dx = \int {\frac{x}{{\sqrt u }}} \left( {\frac{1}{{4x}}} \right)du \cr
& = \frac{1}{4}\int {\frac{1}{{\sqrt u }}} du \cr
& = \frac{1}{4}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = \frac{2}{4}\sqrt u + C \cr
& = \frac{1}{2}\sqrt u + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = 1 + 2{x^2} \cr
& = \frac{1}{2}\sqrt {1 + 2{x^2}} + C \cr
& {\text{Then,}} \cr
& \int_0^2 {\frac{x}{{\sqrt {1 + 2{x^2}} }}} dx = \left[ {\frac{1}{2}\sqrt {1 + 2{x^2}} } \right]_0^2 \cr
& \left[ {\frac{1}{2}\sqrt {1 + 2{x^2}} } \right]_0^2 = \frac{1}{2}\left[ {\sqrt {1 + 2{x^2}} } \right]_0^2 \cr
& = \frac{1}{2}\left[ {\sqrt {1 + 2{{\left( 2 \right)}^2}} } \right] - \frac{1}{2}\left[ {\sqrt {1 + 2{{\left( 0 \right)}^2}} } \right] \cr
& = \frac{1}{2}\sqrt 9 - \frac{1}{2}\sqrt 1 \cr
& = 1 \cr} $$
