Mathematics

If $$\displaystyle\int x^5e^{-x^2}dx=g(x)\cdot e^{-x^2}+C$$ then the value of $$g(-1)$$ is?


ANSWER

$$-\dfrac{5}{2}$$


SOLUTION
Put $$x^2=t$$
$$2xdx=dt$$
$$\displaystyle\int t^2e^{-t}\dfrac{dt}{2}$$
$$=\dfrac{1}{2}[-t^2\cdot e^{-t}+2\displaystyle\int te^{-1}dt]+c$$
$$=\dfrac{1}{2}[-t^2\cdot e^{-t}-2te^{-t}+\displaystyle\int 2e^{-t}dt]+c$$
$$=\dfrac{1}{2}(-t^2e^{-t}-2(te^{-1}+e^{-t}))+c$$
$$=\dfrac{-(x^4+2x^2+2)e^{-x^2}}{2}+c$$
$$g(x)=\dfrac{-(x^4+2x^2+2)}{2}$$
$$g(-1)=-\dfrac{5}{2}$$.
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