Answer
$$\frac{3}{2}\ln \left( {2 + {x^2}} \right) - \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{3x - 1}}{{2 + {x^2}}}} dx \cr
& {\text{write in terms of }}u.{\text{ let }}u = x,\,\,\,du = dx,\, \cr
& \int {\frac{{3x - 1}}{{2 + {x^2}}}} dx = \int {\frac{{3u - 1}}{{2 + {u^2}}}} du \cr
& = \int {\frac{{3u - 1}}{{2 + {u^2}}}} du \cr
& = \int {\frac{{3u - 1}}{{{{\left( {\sqrt 2 } \right)}^2} + {u^2}}}} du \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral by the formula 71}} \cr
& \left( {71} \right)\int {\frac{{bu + c}}{{{a^2} + {u^2}}}} du = \frac{b}{2}\ln \left( {{a^2} + {u^2}} \right) + \frac{c}{a}{\tan ^{ - 1}}\frac{u}{a} + C \cr
& {\text{with }}a = \sqrt 2 ,\,\,\,b = 3{\text{ and }}c = - 1 \cr
& \int {\frac{{3u - 1}}{{2 + {u^2}}}} du = \frac{3}{2}\ln \left( {{{\left( {\sqrt 2 } \right)}^2} + {u^2}} \right) + \frac{{ - 1}}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{u}{{\sqrt 2 }} + C \cr
& = \frac{3}{2}\ln \left( {2 + {u^2}} \right) - \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{u}{{\sqrt 2 }} + C \cr
& \cr
& {\text{write in terms of }}x,{\text{ replace }}x{\text{ for }}u \cr
& = \frac{3}{2}\ln \left( {2 + {x^2}} \right) - \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{x}{{\sqrt 2 }} + C \cr} $$
