If $\int x^{5} e^{-x^{2}} d x=g(x) e^{-x^{2}}+c$, where $c$ is $a$ constant of integration, then $g(-1)$ is equal to :
Correct Option: 1
Let $x^{2}=t$ $2 x d x=d t$
$\Rightarrow \frac{1}{2} \int \mathrm{t}^{2} \cdot \mathrm{e}^{-\mathrm{t}} \mathrm{dt}=\frac{1}{2}\left[-\mathrm{t}^{2} \cdot \mathrm{e}^{-\mathrm{t}}+\int 2 \mathrm{t} \cdot \mathrm{e}^{-\mathrm{t}} \cdot \mathrm{dt}\right]$
$=\frac{1}{2}\left(-t^{2} \cdot e^{-t}\right)+\left(-t \cdot e^{-t}+\int 1 \cdot e^{-t} \cdot d t\right)$
$=-\frac{t^{2} e^{-t}}{2}-t e^{-t}-e^{-t}=\left(-\frac{t^{2}}{2}-t-1\right) e^{-t}$
$=\left(-\frac{x^{4}}{2}-x^{2}-1\right) e^{-x^{2}}+C$
$g(x)=-1-x^{2}-\frac{x^{4}}{2}+k e^{x^{2}}$
for $k=0$
$g(-1)=-1-1-\frac{1}{2}=-\frac{5}{2}$
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