Determine if $ \int_3^{\infty}\frac{x+1}{\sqrt{x^4-x}}\,dx $ converges Determine whether the following integral is convergent or divergent without evaluating it.
￼￼￼(Whichever answer is correct, you must show why it is true.)
$$
\int_3^{\infty}\frac{x+1}{\sqrt{x^4-x}}\,dx
$$

I know that convergence depends on the existence of the limit. How do you find the limit of this equation? Is there a short cut (theorem, principle, memorizable equation) that simplifies this?

This is the first step...

$$\lim\limits_{t \to \infty} \int_3^t\frac{x+1}{\sqrt{x^4-x}}$$
\begin{equation*}
\int_3^{\infty}\frac{x+1}{\sqrt{x^4-x}}\,dx \;&gt;\;
\int_3^{\infty}\frac{x+1}{\sqrt{x^4}}\,dx  \;=\;
\int_3^{\infty}\frac{x+1}{x^2}\,dx \;&gt;\;
\int_3^{\infty}\frac{x}{x^2}\,dx \;=\;
\int_3^{\infty}\frac{1}{x}\,dx
\end{equation*}
The last integral diverges.
HINT:

$\displaystyle \frac{1}{2^{x}} &lt; \frac{(x+1)}{\sqrt{x^4 - x}}$

Does $\displaystyle \int_{3}^{\infty} 2^{-x} dx$ converge?

Then find a function greater.