\(\displaystyle f(x) \ = \ \int 4te^{[(-2t)^{2}]}dt \ = \ 4\int te^{4t^{2}}dt\)
\(\displaystyle Let \ u \ = \ 4t^{2}, \ then \ du \ = \ 8tdt\)
\(\displaystyle Hence, \ \frac{1}{2}\int e^{u}du \ = \ \frac{e^{u}}{2} + \ C \ = \ \frac{e^{4t^{2}}}{2}+ \ C\)
