Limit Comparison Test for Improper Inegrals Given the integral :
\begin{equation}
\int_{1}^6 \frac{1}{x^p-x}\;dx
\end{equation}
I'm trying to find for which values of p the given integral converges. 
I know that there is a singularity point at $x=0$, and I'm trying to use the limit comparison test but I'm struggling to find the function that I should use for comparison. 
I was thinking about the concept "the weak wins at zero, the strong wins at infinity" and tried to apply it to x=1. 
I split it to two parts, $p>1$ and $ p<1$:
When $p>1$, I tried to compare with the function $$\frac{1}{x}$$  since $x^0$ is the weakest in numerator and $x^1$ is the weakest in denominator but it didn't work.
Any suggestions? Thanks in advance!
 A: One case of the limit comparison test states that if $f$ and $g$ are positive integrable functions on $[a,b]$ for all $1 < a < b$ and
$$\lim_{x \to 1+} \frac{f(x)}{g(x)} = L \neq 0$$
then the divergence of $\int_1^b g(x) \, dx $ implies the divergence of $\int_1^b f(x) \, dx $.
For $p > 1$ we have using L'Hopital's rule
$$\lim_{x \to 1+} \frac{\frac{1}{x^p-x}}{\frac{1}{x-1}} = \lim_{x \to 1+} \frac{x-1}{x^p-x} =  \lim_{x \to 1+} \frac{1}{px^{p-1}-1}= \frac{1}{p-1}\neq 0$$
Since $\int_1^6 \frac{dx}{x-1}$ diverges it follows by the limit comparison test that $\int_1^6 \frac{dx}{x^p-x}$ diverges.
For $0 < p < 1$, the integrand is negative so we can repeat the argument and apply the limit comparison test for the integral $\int_1^6 \frac{dx}{x-x^p}$. In this case,
$$\lim_{x \to 1+} \frac{\frac{1}{x-x^p}}{\frac{1}{x-1}} = \lim_{x \to 1+} \frac{x-1}{x-x^p} =  \lim_{x \to 1+} \frac{1}{1-px^{p-1}}= \frac{1}{1-p}\neq 0$$
and we see that  $\int_1^6 \frac{dx}{x-x^p}$ diverges (to $+\infty$ as the integrand is nonnegative).  Hence, $\int_1^6 \frac{dx}{x^p-x}$ diverges to $-\infty$.
