$\cfrac{{3x}}{{1 + 2{x^4}}}$

: Let $I = \int {\cfrac{{3x}}{{1 + 2{x^4}}}dx}$

Put ${x^2} = t$ $\Rightarrow$ $2xdx = dt$ $\Rightarrow$ $xdx = \cfrac{{dt}}{2}$

$\therefore$ $I = \cfrac{3}{2}\int {\cfrac{{dt}}{{1 + 2{t^2}}}} = \cfrac{3}{4}\int {\cfrac{{dt}}{{\cfrac{1}{2} + {t^2}}} = \cfrac{3}{4}\int {\cfrac{{dt}}{{{{\left( {\cfrac{1}{{\sqrt 2 }}} \right)}^2} + {t^2}}}} }$

$= \cfrac{3}{4} \cdot \cfrac{1}{{1/\sqrt 2 }}{\tan ^{ - 1}}\left( {\cfrac{t}{{1/\sqrt 2 }}} \right) + C = \cfrac{3}{{2\sqrt 2 }}{\tan ^{ - 1}}\left( {\sqrt 2 t} \right) + C$

$= \cfrac{3}{{2\sqrt 2 }}{\tan ^{ - 1}}\left( {\sqrt 2 {x^2}} \right) + C$