Answer
$$\frac{3}{2}\ln 5$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\frac{{3x}}{{1 + {x^2}}}} dx \cr
& {\text{Set }}u = 1 + {x^2} \to du = 2xdx,\,\,dx = \frac{{du}}{{2x}} \cr
& u = 1 + {x^2},{\text{ for }}x = 0,\,\,u = 1 \cr
& u = 1 + {x^2},{\text{ for }}x = 2,\,\,u = 5 \cr
& {\text{using the substitution}} \cr
& \int_0^2 {\frac{{3x}}{{1 + {x^2}}}} dx = \int_1^5 {\frac{{3x}}{u}} \left( {\frac{{du}}{{2x}}} \right) \cr
& = \frac{3}{2}\int_1^5 {\frac{{du}}{u}} \cr
& {\text{integrate}} \cr
& = \frac{3}{2}\left[ {\ln \left| u \right|} \right]_1^5 \cr
& {\text{evaluate}} \cr
& = \frac{3}{2}\left[ {\ln \left| 5 \right| - \ln \left| 1 \right|} \right] \cr
& = \frac{3}{2}\ln 5 \cr} $$
