Evaluate
$\displaystyle \int\limits_{-1}^{1}\frac{3+sin^3 x}{1+x^2}dx$
Note that $\displaystyle \int\limits_{-1}^{1}\frac{\sin^3 x}{1+x^2} \, dx = 0$ since $\displaystyle \sin^3 x$ is an odd function.
So $\displaystyle \int\limits_{-1}^{1}\frac{3 - \sin^3 x}{1+x^2} \, dx = \int\limits_{-1}^{1}\frac{3}{1+x^2} \, dx$ which you should be able to easily do.
He didn't say
$\displaystyle
\frac{sin^3x}{1+x^2} = 0. $ (note the number 2)
He said that
$\displaystyle
\int_{-1}^1 \frac{sin^3x}{1+x^2}\, dx = 0.
$
In fact, if $\displaystyle f(x)$ is odd and continuous on $\displaystyle [-a,a]$ then
$\displaystyle \int_{-a}^a f(x) \, dx = 0.$