Comparison theorem of $\int_{2}^{4}\frac{x}{\sqrt{x-1}\sqrt{x-2}}dx$

Question

Determine if $$\int_{2}^{4}\frac{x}{\sqrt{x-1}\sqrt{x-2}}dx$$ converges or diverges.

When I attempted this question I thought it diverged since it has a VA at x=2.

By bounding it below and got $$\frac{1}{\sqrt{x}\sqrt{x-2}}$$

However it turns out this converges. I am trying to prove this using comparison theorem. Could I get some help on how to bound this from above?

Thanks.

Answer 1

$\frac x {\sqrt {x-1}\sqrt {x-2}} \leq \frac 4 {\sqrt 1 \sqrt {x-2}}$ and $\int_2^{4} \frac 1 {\sqrt {x-2}}dx=2(x-2)^{1/2}|_2^{4}=2\sqrt 2$. Hence the integral converges.