### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

Evaluate: $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Solution: $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\infty}e^{-x^2} dx +2 \int_{0}^{\infty}(2x^4)e^{-x^2} dx = \int_{-\infty}^{\infty}e^{-x^2} dx +4\int_{0}^{\infty}x^4e^{-x^2} dx .$$ It is well-known that $$\int_{-\infty}^{\infty}e^{-x^2} dx = \sqrt{\pi}.$$
We will calculate the second integral $$4\int_{0}^{\infty}x^4e^{-x^2}dx.$$
Substitute $u = x^2$ then the integral transforms to $2 \int_{0}^{\infty}u^{\frac{3}{2}}e^{-u}du = 2 \Gamma(\frac{3}{2}) = 2 \cdot \frac{3}{4}\cdot \sqrt{\pi} = \frac{3}{2} \sqrt{\pi}$
Hence, $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \sqrt{\pi} + \frac{3}{2} \sqrt{\pi} = \frac{5}{2} \sqrt{\pi}.$$

1. Sir how the integral write into two part

2. $x^4$ is an even function. So -a to a integral even function = 2 times 0 to a integral of the function. is it clear? I have edited the answer. plz check. also plz follow me be mail and subscribe.

3. I have updated the solution. Hopefully now its clear. Any doubts, feel free to ask. All the best.

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### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...