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.