Why the definite integral of the function x^2*sin2x/(x^2+1) is 0? x=-3 to x=3

The answer is very simple one. Because the given function
is continuous and odd.

But let's see how it works. We know
that if a function is continuous over a range [-a ; a], then the following identity is
true:

Int f(x)dx(-a --> a) = Int [f(x) + f(-x)]dx (
0--> a)

Furthermore, if the function is odd, then
the above identity is changing in:

Int f(x)dx(-a -->
a) = 0

A function is odd if and only if f(-x) =
-f(x).

We'll verify if the given function is odd. We'll
substitute x by -x:

f(-x) =
(-x)^2*sin2(-x)/[(-x)^2+1]

f(-x) = x^2*sin (-2x)/(x^2 +
1)

Since sine function is odd, then sin (-2x) = - sin
2x.

f(-x)= -x^2*sin (2x)/(x^2 +
1)

f(-x) =
-f(x)

So, the given function
is odd, then :

Int
[x^2*sin2x/(x^2+1)]dx (-3 --> 3) = 0