Answer
Divergent
Work Step by Step
Define $f(x)=\frac{x+1}{\sqrt{x^4-x}}$ and $g(x)=\frac{1}{x}+\frac{1}{x^2}$.
For $x\geq 1$,
$-x<0$
$x^4-x\frac{x+1}{x^2}$
$f(x)>g(x)$
Find the convergence for $\int_1^\infty g(x) dx$:
$\int_1^\infty \frac{1}{x}+\frac{1}{x^2} dx=[\ln x-\frac{1}{x}]_1^\infty=(\ln \infty -\frac{1}{\infty})-(\ln 1-\frac{1}{1}=(\infty -0)-(0-1)=\infty (\text{divergent})$
Since $f(x)>g(x)$ for $x\geq 1$ and $\int_1^\infty g(x)$ is divergent, using the Comparison Theorem it concludes that $\int_1^\infty \frac{x+1}{\sqrt{x^4-x}}dx$ is divergent.
