TITLE: How do I solve $\int_{-5}^{5} \frac{x^3 \sin^2x}{x^4 +2x^2+1}\,dx$?
QUESTION [6 upvotes]: $$\int_{-5}^{5} \frac{x^3 \sin^2x}{x^4 +2x^2+1}\,dx$$
I only have basic calculus but this is what I tried.
Firstly this can't be integrated directly.
I tried to make a substitution, letting $u$ equal to varies parts of this expression but to no avail.
I tried integration by parts but it just got too messy.
I did notice that $x^4+2x^2+1 = (x^2+1)^2$ but that didn't get me anywhere either.
I did plot this in Geogebra and noticed that this function is origin-symmetric so $2\int_{0}^{5}f(x)\,dx$ could be used to simplify things after the integration but it doesn't help to do the actual integration. 
How does one go about tackling this particular integral?

REPLY [2 votes]: Let $f(x)=\frac{x^3\sin^2x}{(x^2+1)^2}$.
Since $f(-x)=-f(x)$, we obtain:
$$\int_{-5}^5f(x)dx=\int_{-5}^0f(x)dx+\int_{0}^5f(x)dx=$$
$$=\int_{-5}^0\left(-f(-x)\right)dx+\int_{0}^5f(x)dx=$$
$$=\int_{0}^5f(-x)dx+\int_{0}^5f(x)dx=$$
$$=\int_{0}^5\left(f(-x)+f(x)\right)dx=0.$$