evaluate $\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$ I have to evaluate 

$$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

solution i tried-The given function in the integral i.e $(1+2x^4)e^{-x^2}$ is a even function so by property of even function we can write it as 

$$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx=2\int_{0}^{\infty}(1+2x^4)e^{-x^2} dx$$
$$\implies2\int_{0}^{\infty}e^{-x^2} dx+4\int_{0}^{\infty}x^4e^{-x^2} dx\;\;\;\;\;\;\;\;\;\;\;...(1)$$
  we know that value of $\displaystyle\int_0^{\infty}e^{-x^2}=\frac{\sqrt \pi}{2}$ 

so from this (1) will become

$$2\times\frac{\sqrt\pi}{2}+4\int_0^{\infty}x^4e^{-x^2}dx$$

The doubt here is that i have n idea how to solve the second part ,I tried to put $x^2=t$ but still not get the answer please provide me a hint so that i can solve it further 
Thank you
 A: Using the substitution $t=x^2$ does work:
$$2\int_{0}^\infty x^4 \exp(-x^2)\,\mathrm dx=\int_0^\infty t^{\frac32}\exp(-t)\,\mathrm dt=\Gamma\left(\frac52\right)=\frac{3\sqrt\pi}4,$$
see particular values of the Gamma function or Gamma function of positive half-integer.
A: Since $\int_{\Bbb R}e^{-ax^2}dx=\sqrt{\pi}a^{-1/2}$, applying $\partial_a^2$ gives $\int_{\Bbb R}x^4e^{-ax^2}dx=\frac34\sqrt{\pi}a^{-5/2}$. Now take $a=1$.
A: $I_n=\int_{0}^{\infty}x^n.e^{-x^2}dx$ $\Rightarrow I_n=\frac{1}{2}\int_{0}^{\infty}x^{n-1}.2xe^{-x^2}dx$
Using by parts we get $I_n=\frac{1}{2}x^{n-1}.\frac{e^{-x^2}}{-1}|^\infty_0-\frac{1}{2}\int_{0}^{\infty}(n-1)x^{n-2}.\frac{e^{-x^2}}{-1}dx$
$I_n=0+\frac{n-1}{2}I_{n-2}$
$I_4=\frac{4-1}{2}\frac{2-1}{2}.I_0$
$I_4=\frac{3}{8}\sqrt\pi$
A: Note that $$I_n=\int_{-\infty}^{\infty} x^{2n} e^{-x^2} dx=\Gamma(n+1/2)$$
So yhe Given integgral is $$I=I_0+2I_2=[\Gamma(1/2)+2 \Gamma(1/2+2)]=[\sqrt{\pi}
+2\Gamma(5/2)]=\sqrt{\pi}+(3/2)\sqrt{\pi} =\frac{5 \sqrt{\pi}}{2}$$
A: Hint:
By parts,
$$\int x^4e^{-x^2}dx=\frac12\int x^3\,2xe^{-x^2}dx=-\frac12x^3e^{-x^2}+\frac12\int6x^2e^{-x^2}.$$
Integrating by parts a second time, you will get rid of the factor $x^2$.

Alternatively,
$$(P(x)e^{-x^2})'=(P'(x)-2xP(x))e^{-x^2}$$
and you can solve
$$P'(x)-2xP(x)=1+2x^4$$ where the solution will be a cubic polynomial.

$$(3ax^2+2bx+c)-2x(ax^3+bx^2+cx+d)=1+2x^4\to P(x)=-x^3-\frac{3x}2.$$

Note that an extra term is needed.
