# Determine whether the integral is convergent or divergent. $\int^{\infty}_{-\infty} 43\frac{x^2}{9+x^6}dx$

Determine whether the integral is convergent or divergent. $\int^{\infty}_{-\infty} 43\frac{x^2}{9+x^6}dx$

My answer is $\frac{43\pi}{9}$

Yet, I am unsure if it is convergent or divergent since I believe $\frac{43\pi}{9}$ is not a real number. However, I am not sure. The $\pi$ is throwing me off?

• $\dfrac{43\pi}{9}\approx 15.0098$ is an irrational number, and it is also definitely a real number. EDIT: Besides that, your answer seems to be correct. – Corellian Oct 25 '15 at 23:54
• Real numbers are closed under multiplication; since $\pi$ is real and $43/9$ is rational (hence real), their product is real. – Clayton Oct 25 '15 at 23:55
• Where did the "43" come from? It seems to be irrelevant. – marty cohen Oct 26 '15 at 0:15

If the integral evaluates in some way to $43\pi/9$, then in some sense it converges.
You're integrating an everywhere non-negative function. When you do that, "convergence" is the same as saying the value is less than $\infty$. Notice that $\dfrac{x^2}{9+x^6} \le \dfrac{x^2}{x^6} = \dfrac 1 {x^4}$ and therefore the integral from $a$ to $\infty$ is finite whenever $a$ is a strictly positive number.
But $\dfrac 1 {x^4}$ has a vertical asymptote at $0$, so we look at $\dfrac{x^2}{9+x^6}$ and see that, since its denominator is always strictly positive, it is everywhere continuous, and hence bounded on the interval from $0$ to the aforementioned positive number $a$. Thus we have $\displaystyle \int\cdots <\infty$. It converges.