$\cfrac{x}{{{e^{{x^2}}}}}$

: Let $I = \int {\cfrac{x}{{{e^{{x^2}}}}}dx}$
Put ${x^2} = t$ $\Rightarrow$ $2x\,dx = dt$

$\therefore$ $I = \cfrac{1}{2}\int {\cfrac{{dt}}{{{e^t}}} = \cfrac{1}{2}\int {{e^{ - t}}dt} = \cfrac{1}{2}\left( {\cfrac{{{e^{ - t}}}}{{ - 1}}} \right)} + C = - \cfrac{1}{{2{e^t}}} + C$

$= - \cfrac{1}{{2{e^{{x^2}}}}} + C$