since the interval of the integral is symmetric about the y - axis
together with the integrand x/(x² + 9), which is an odd function
that means f(- x) = - f(x), it's rotational symmetric about the origin
for the interval [- a，0], the value of the bounded area is negative(or positive)
for the interval [0, a], the value of the bounded area is positive(or negative)
as a result, the areas cancels each other and the total value of the integral is zero.
∫(- a→a) ƒ(x) dx = 0 if ƒ(x) is an odd function
on the contrary, if the integrand is an even function, i.e. f(- x) = f(x)
this function is symmetric about the y - axis
which means the two areas bounded in the two intevals are totally identical
therefore we have ∫(- a→a) ƒ(x) dx = 2∫(0→a) ƒ(x) dx
i.e. 1/(x² + 9) is an even function
