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.
On
Your inequality does not hold because the left handside is unbounded but the right handside is bounded on the interval $[5,6]$.
Actually you have $ \frac{1}{(x-3)\sqrt {x-5}}$ $\le \frac{1}{2\sqrt{x-5}}$ Then substitute $y=x-5$. Now, you obtain $$I\leq \int_5^6 \dfrac{dx}{2\sqrt{x-5}}=\frac12\int_0^1\dfrac{dy}{y^{1/2}}$$
Here the right hand side converges by the p-test, so it implies the convergence of your original integral $I$.
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*}