$\int\limits^{5}_{-5}(3x^5+\sqrt{5}x^3).e^{x^2}dx$ is equal to :

0

Given integral:

$\int_{-5}^{5}\left(3x^5+\sqrt{5}\,x^3\right)e^{x^2}\,dx$

$e^{x^2}$ is an even function, while $3x^5$ and $x^3$ are odd functions.

Product of an odd function and an even function is an odd function.

The integral of an odd function over symmetric limits $[-a,a]$ is zero.

final answer: The value of the integral is 0