The integrand is odd.
[tex]f(x) = \dfrac{x^3 + 4x}{x^4 + 5} \implies f(-x) = \dfrac{(-x)^3 + 4(-x)}{(-x)^4 + 5} = \dfrac{-x^3-4}{x^4+5} = -f(x)[/tex]
The integral of an odd function over a symmetric interval is zero, so
[tex]\displaystyle \int_{-4}^4 \frac{x^3+3x}{x^4+5} \, dx = 0[/tex]
This is because
[tex]\displaystyle \int_{-a}^a f(x) \, dx = \int_{-a}^0 f(x) \, dx + \int_0^a f(x) \, dx \\\\ ~~~~ = -\int_0^{-a} f(x) \, dx + \int_0^a f(x) \, dx \\\\ ~~~~ = \int_0^{-a} f(-x) \, dx + \int_0^a f(x) \, dx \\\\ ~~~~ = -\int_0^a f(x) \, dx + \int_0^a f(x) \, dx = 0[/tex]
where we substitute [tex]x\mapsto-x[/tex] in the second-to-last equality.
