Change of variables $s^2=r^2+b^2$

Question

I am trying to understand why this change of variables breaks down.

Let $f$ be a smooth continuous function on an interval $[0,a]$ with $a>0$, and let $b>0$.

I want to compute $$\int_0^af(r^2+b^2)\,dr.$$

I want to use the change of variables $s^2=r^2+b^2$, $2s\,ds=2r\,dr$ so $dr=\frac{s}{\sqrt{s^2-b^2}}ds.$ Therefore $$\int_0^af(r^2+b^2)\,dr=\int_{b}^{\sqrt{a^2+b^2}}f(s^2)\frac{s}{\sqrt{s^2-b^2}}\,ds.$$ But this integral does not converge as $s\rightarrow b^+$. I am confused because a smooth continuous function on a closed interval should be finite.