I have to analyze the character of the following improper integral:
$$\int_{1}^{3}\frac{dx}{\sqrt{(-x^3+x^2+9x-9)^{1/2}}}$$ I don't really know how to analyze it.
I have decomposed the denominator as: $[(x+3)(x-3)(x-1)]^{1/4}$. I am not sure how I would find the answer to this question. I am trying to use the limit comparison test or the limit criterion, but I don't see how I would get the answer.
For $1<x\le2$, $5\le(3-x)(3+x)<8$ and $$\int_1^2\frac{dx}{(x-1)^{1/4}}=\left.\frac43(x-1)^{3/4}\right|_1^2=\frac43$$ So $$\begin{align}\frac438^{-1/4}&=8^{-1/4}\int_1^2\frac{dx}{(x-1)^{1/4}}\le\int_1^2\frac{dx}{\left((3-x)(x+3)(x-1)\right)^{1/4}}\\ &\le5^{-1/4}\int_1^2\frac{dx}{(x-1)^{1/4}}=\frac435^{-1/4}\end{align}$$ Similarly you can show that $$\frac4312^{-1/4}\le\int_2^3\frac{dx}{\left((3-x)(x+3)(x-1)\right)^{1/4}}\le\frac435^{-1/4}$$ So we may conclude that $$\frac43\left(8^{-1/4}+12^{-1/4}\right)\le\int_1^3\frac{dx}{\left((3-x)(x+3)(x-1)\right)^{1/4}}\le\frac83\left(5^{-1/4}\right)$$ It converges because is is bounded by convergent integrals.