Qs: Use the limit comparison test to determine whether the following integrals converge
$$\int_{x=0}^\infty \frac{x^\frac{1}{3}+5}{x^2-6x+10}$$
I did the question and found that it diverges, but it turns out this integral converges by the Limit Comparison Test.
My Answer: f(x) = (the equation above) and g(x) = $\frac{1}{x^\frac{5}{3}}$
$$\lim_{x\to \infty} f(x)/g(x) = 1$$ which is positive and finite.
$$\int_{x=0}^\infty \frac{1}{x^\frac{5}{3}} diverges$$
So by the LCT, my original integral diverges
However, this integral converges by the LCT and I just don't see how. Can anyone help?
The trick is that, look at the neighborhood of $x=0$, the integrand is continuous, so just split it: \begin{align*} \int_{0}^{\infty}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx=\int_{0}^{1}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx+\int_{1}^{\infty}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx, \end{align*} and \begin{align*} \int_{0}^{1}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx \end{align*} exists by the continuity of $\dfrac{x^{1/3}+5}{x^{2}-6x+10}$ on the compact set $[0,1]$.
Now use Limit Comparison Test to $1/x^{5/3}$ on $[1,\infty)$ to conclude that \begin{align*} \int_{1}^{\infty}\dfrac{x^{1/3}+5}{x^{2}-6x+10}dx \end{align*} exists.