How to determine whether an integral is convergent or divergent? For this question, I'm not sure if I'm doing it right, can anyone please help me out?
Determine whether the following integral is convergent or divergent.
$$\int_5^6 \frac 1 {(x-3)\sqrt {x-5}} \, dx$$
$$ \frac{1}{(x-3)\sqrt {x-5}}\le \frac{1}{x-3}$$
Since $\int_5^6 \frac 1 {x-3} \,dx$ converges, $\int_5^6 \frac 1 {(x-3)\sqrt {x-5}} \, dx$ must also converge.

The inequality is not valid. Rather, 
\begin{align*}
0<\dfrac{1}{(x-3)\sqrt{x-5}}<\dfrac{1}{2\sqrt{x-5}},~~~~x\in(5,6],
\end{align*}
and 
\begin{align*}
\int_{5}^{6}\dfrac{1}{\sqrt{x-5}}&=\lim_{\eta\rightarrow 5^{+}}\int_{\eta}^{6}\dfrac{1}{\sqrt{x-5}}dx\\
&=\lim_{\eta\rightarrow 5^{+}}2\sqrt{x-5}\bigg|_{\eta}^{6}\\
&=\lim_{\eta\rightarrow 5^{+}}\left(2-2\sqrt{\eta-5}\right)\\
&=2\\
&<\infty.
\end{align*}
