Why does $\int_{0}^{0.5}\frac{1}{x^2-0.1}$ does't converge and $\int_{0}^{0.5}\frac{1}{x^2-0.3}$ does? In order to prove that $\int_{0}^{0.5}\frac{1}{x^2-1}$ converges I compared it to $\int_{0}^{0.5}\frac{1}{x}$ which converges, by checking that $\lim \frac{\frac{1}{x^2-1}}{\frac{1}{x}}=0$ when $x \to 0$. Then I went to Wolfram Alpha  and tried to check $\int_{0}^{0.5}\frac{1}{x^2-0.1}$ (It is suppose to converges by the same test) but it said that it diverges, while $\int_{0}^{0.5}\frac{1}{x^2-0.3}$ doesn't. (with $0.2$ it just couldn't compute). What's really going on there between $0.1$ to $0.3$? Is W.A wrong?
Edit: Sorry! $\int_{0}^{0.5}\frac{1}{x}$ obviously doesn't converge. So, in addition to the rest of the question, how can I prove my original integral does converge?
Thank you very much.
 A: Between $x=0$ and $x=0.5$, the function $\frac{1}{x^2-1}$ is perfectly respectable! Note that the denominator is never $0$ in our interval.  The largest absolute value is reached at $x=0.5$.  So your function has no issues, it is continuous on a closed interval.  For the problem you were initially considering, we are finished.
But the question you were led to ask is more interesting, and shows a good effort to understand the situation.
Look first at $\frac{1}{x^2-0.3}$. The denominator is $0$ at $x=\pm\sqrt{0.3}$. The positive root is roughly $0.547722$, outside our interval, though not by much. Thus the function $\frac{1}{x^2-0.3}$ is well-behaved in the interval $[0,0.5]$.
Look now at $\frac{1}{x^2-0.1}$.  The denominator is $0$ at $x=\pm\sqrt{0.1}$. The positive root is about $0.3162278$, and this is inside our interval. So our function blows up inside our interval, and there may be a problem. Indeed there is.
You know that a function can blow up, but despite that the integral converges. A standard example is $\int_0^1 \frac{dx}{\sqrt{x}}$.  We will show that $\int_0^{0.5}\frac{dx}{x^2-0.1}$ diverges. 
As mentioned above, there is potential trouble at $\sqrt{0.1}$. To make typing easier, let $a=\sqrt{0.1}$.  Our function is not defined at $x=a$, and blows up near $x=a$. Recall that $a$ is in our interval.  When we are dealing with a singularity inside our interval, it is useful to break up the interval into two integrals, in this case from $0$ to $a$ and from $a$ to $0.5$.
We will show that $\int_0^a \frac{dx}{x^2-0.1}$ diverges. (The integral from $a$ to $0.5$ also does, but showing that one of the integrals is bad is enough.)
So we are looking at the integral $\int_0^a\frac{dx}{x^2-a^2}$.
For no good reason, except for a preference for the positive, we look instead at
$$\int_0^a \frac{dx}{a^2-x^2}.$$
Make the change of variable $w=a-x$. Note that $a^2-x^2=(a-x)(a+x)=w(2a-w)$. Quickly we arrive at
$$\int_{w=0}^a \frac{dw}{w(2a-w)}.$$
This integral diverges, by comparison with $\int_0^a\frac{dw}{w}$, which, as pointed out by anonymous, diverges.   
A: You have a mistake. The integral $\int_{0}^{0.5} dx/x$ does not converge! in fact, we can easily calculate it, as the antiderivative of $1/x$ is $lnx$, and $\lim_{x\to 0^+} ln(x) = -\infty$.
