I'm trying to solve the integral below. I'm not getting the right answer no matter what. Can you tell me why my method is wrong? I'm applying the rule for integrating $x^n$ (i.e. $\smash{\frac{x^{n+1}}{n+1}}$). Why it's not working? (ditch $\vec{\mathbf{a}}_\phi$ if it confuses you)
$$\large{\begin{align} \vec{\mathbf{B}}&=\frac{\mu_0I\rho}{4\pi}\int_a^b\frac{\mathrm{d}z}{[\rho^2+z^2]^{3/2}}\,\vec{\mathbf{a}}_\phi \\ &=\frac{\mu_0I\rho}{4\pi}\left[\frac b{\sqrt{\rho^2+b^2}}-\frac a{\sqrt{\vphantom{b^2}\rho^2+a^2}}\right]\vec{\mathbf{a}}_\phi \end{align}}$$
My work:
\begin{align} \vec{\mathbf{B}}&=\frac{\mu_0}{4\pi}\int_a^b\frac{I\rho\,\mathrm{d}z}{(z^2+\rho^2)^{3/2}}\,\vec{\mathbf{a}}_\phi \\[10pt] &=\frac{\mu_0I\rho}{4\pi}\int_a^b\frac{\mathrm{d}z}{(z^2+\rho^2)^{3/2}}\,\vec{\mathbf{a}}_\phi \\[10pt] &=\frac{-2\mu_0I\rho}{4\pi}\bigg[(z^2+\rho^2)^{-1/2}\bigg]_a^b\vec{\mathbf{a}}_\phi \\[10pt] &=\frac{\mu_0I\rho}{2\pi}\left(\frac1{\sqrt{b^2+\rho^2}}-\frac1{\sqrt{a^2+\rho^2}}\right)\vec{\mathbf{a}}_\phi \end{align}
Hint: $$\int \frac{1}{(x^2+\alpha^2)^{3/2}} dx= \frac{x}{\alpha^2\sqrt{x^2+\alpha^2}}+c$$