Let $I=\int x^5{e}^{-x^2}dx$

Let, $-x^2=t\Rightarrow xdx=-\frac{1}{2}dt$

$\therefore I=-\frac{1}{2}\int t^2{e}^{t}dt$

By using integrating by parts, i.e. $\int \left(u·v\right)dx=u\int vdx-\int \left[\frac{d}{dx}u\int vdx\right]dx+c,$ we get

$I=-\frac{1}{2}\left[t^2\int e^{t}dt-\int \left(2t\int e^{t}dt\right)dt\right]+c$

$\Rightarrow I=-\frac{1}{2}\left[t^2{e}^t-2\int t{e}^{t}dt\right]+c$

Again, applying integrating by parts, we get

$\Rightarrow I=-\frac{1}{2}\left[t^2{e}^t-2\left(t\int e^{t}dt-\int 1·e^{t}dt\right)\right]+c$

$\Rightarrow I=-\frac{1}{2}\left[t^2{e}^t-2\left(t{e}^t-e^t\right)\right]+c$

$\Rightarrow I=-\frac{1}{2}\left[t^2{e}^t-2t{e}^t+2e^t\right]+c$

$\Rightarrow I=-\frac{e^t}{2}\left[t^2-2t+2\right]+c$

$\Rightarrow I=-\frac{e^{-x^2}}{2}\left[x^4+2x^2+2\right]+c$

Given, $I=\int x^5{e}^{-x^2}dx=g\left(x\right)e^{-x^2}+c$

$\Rightarrow g\left(x\right)=-\frac{1}{2}\left(x^4+2x^2+2\right)$

$\Rightarrow g\left(-1\right)=-\frac{1}{2}\left({\left(-1\right)}^4+2{\left(-1\right)}^2+2\right)$

$\Rightarrow g\left(-1\right)=-\frac{5}{2}.$