(Alternative solution) Let us rewrite the integrand, so that
∫ 0 ∞ 6 x 3 e − x 2 d x = − 3 ∫ 0 ∞ x 2 ( − 2 x e − x 2 ) d x {\displaystyle \int _{0}^{\infty }6x^{3}e^{-x^{2}}dx=-3\int _{0}^{\infty }x^{2}(-2xe^{-x^{2}})dx}
Use integration by parts where f ′ = − 2 x e − x 2 , g = x 2 , f = e − x 2 , g ′ = 2 x {\displaystyle f^{\prime }=-2xe^{-x^{2}},\,g=x^{2}\,,f=e^{-x^{2}},\,g'=2x} and so
∫ 0 ∞ 6 x 3 e − x 2 d x = − 3 [ x 2 e − x 2 | 0 ∞ − ∫ 0 ∞ 2 x e − x 2 d x {\displaystyle \int _{0}^{\infty }6x^{3}e^{-x^{2}}dx=-3[x^{2}e^{-x^{2}}|_{0}^{\infty }-\int _{0}^{\infty }2xe^{-x^{2}}dx}
= − 3 [ x 2 e − x 2 | 0 ∞ + e − x 2 | 0 ∞ ] {\displaystyle =-3[x^{2}e^{-x^{2}}|_{0}^{\infty }+e^{-x^{2}}|_{0}^{\infty }\,]}
Note that lim x → ∞ x 2 e − x 2 = 0 {\displaystyle \lim _{x\rightarrow \infty }x^{2}e^{-x^{2}}=0} and lim x → ∞ e − x 2 = 0 {\displaystyle \lim _{x\rightarrow \infty }e^{-x^{2}}=0}
Thus, from the first term we have − 3 ( 0 − 0 ) = 0 {\displaystyle -3(0-0)=0} and from the 2nd term we get − 3 ( 0 − 1 ) = 3 {\displaystyle -3(0-1)=3}
Answer: ∫ 0 ∞ 6 x 3 e − x 2 d x = 3 {\displaystyle \color {blue}\int _{0}^{\infty }6x^{3}e^{-x^{2}}dx=3}
and so
![{\displaystyle =-3[x^{2}e^{-x^{2}}|_{0}^{\infty }+e^{-x^{2}}|_{0}^{\infty }\,]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/75b8fd0711798241683a402fdb9ce16baefbe160)
and 
and from the 2nd term we get 
