Why does this integral diverge: $\int_1^{\infty}\frac{x^6}{6x^6 − 1} dx$? $$\int_1^{\infty}\frac{x^6}{6x^6 − 1} dx$$
I would assume it would converge but apparently it diverges. I know it has to do with improper integrals. Can anyone explain? Thank you for your time. 
 A: Since we know that $\dfrac{1}{x^6}>0$ for all $x\geq1$ , we have $\dfrac{1}{6-\dfrac{1}{x^6}}>\dfrac{1}{6}$ and we know that RHS diverges. Hence, by comparison theorem , LHS diverges.
A: The integral turns out to be doubly bad. The behaviour when $x$ is large is the more obvious badness. We show that there is also fatal badness at $1$, by showing that $\displaystyle\int_1^2 \dfrac{x^6\,dx}{x^6-1}$ diverges.
Note that $x^6-1=(x-1)(x^5+x^4+x^3+x^2+x+1)$. When $x\ge 1$, each of the terms $x^5,x^4,x^3,x^2,x, 1$ is $\le x^6$. It follows that if $\epsilon\gt 0$, then
$$I_\epsilon=\int_{1+\epsilon}^2 \dfrac{x^6\,dx}{x^6-1}\gt \int_{1+\epsilon}^2 \frac{1}{6}\cdot \frac{dx}{x-1}.$$
The change of variable $u=x-1$ shows that
$$I_\epsilon\gt \frac{1}{6}\int_\epsilon^1\frac{du}{u}.$$
But it is a familiar fact that $\displaystyle\int_\epsilon^1\dfrac{du}{u}$ blows up as $\epsilon\to 0^+$. 
A: $\frac{x^6}{6x^6-1}$ converges toward 1/6 $(n\rightarrow\infty)$. Therefore, there exists $x_0\in\mathbb{R}$ for that $\frac{x^6}{6x^6-1}\gt\frac{1}{7}\; (x\gt x_0)$
And thus, the $\int_1^\infty\frac{x^6}{6x^6-1}\ge\int_{1}^{x_0} \frac{x^6}{6x^6-1}\text{d}x+\int_{x_0}^\infty \frac{1}{7}\text{d}x$. (This obviously diverges)
A: The reason why this integral diverges is the following. For large $x$ the fraction reaches a constant limit.
\begin{align}
\lim_{x\rightarrow \infty} \frac{x^6}{6x^6-1} = \frac{1}{6}
\end{align}
That means that we integrate a functions that asymptotically behaves like $f(x)\equiv \frac{1}{6}$ which has a diverging integral.
See also Wolfram Alpha for a plot of the integrand.
Concerning  Brian M. Scott regards. He is right. The more correct argumentation would be:
\begin{align}
\frac{x^6}{6x^6-1} \geq \frac{x^6}{6x^6}=\frac{1}{6} \, \forall x \geq 1
\end{align}
So
\begin{align}
\int_1^{\infty}\frac{x^6}{6x^6-1} \, dx \geq \int_1^{\infty}\frac{1}{6} \, dx = \infty
\end{align}
A: I really hate direct comparison test. I always try to use limit comparison, ratio test, etc.


*

*Integral comparison test
$$\int_1^{\infty}\frac{x^6}{6x^6 − 1} dx \not < \infty \iff \sum_{x=1}^{\infty}\frac{x^6}{6x^6 − 1} \not < \infty$$


*Limit comparison test
$$\lim_{x \to \infty}\frac{\frac{x^6}{6x^6 − 1}}{\frac{x^6}{6x^6}} \in (0,\infty)$$


*Test for divergence
$$\lim_{x \to \infty} \frac{x^6}{6x^6} \not = 0$$
Ummm...I regret nothing!
