Partial fractions does it. The partial fraction expansion of the integrand is
$ \frac{1}{4(y-\sqrt{a-1})^2} + \frac{1}{4(y+\sqrt{a-1})^2} + \frac{1}{4 \sqrt{a-1} (y - \sqrt{a-1})} - \frac{1}{4 \sqrt{a-1} (y + \sqrt{a-1})} $
so an antiderivative is
$ -\frac{1}{4(y-\sqrt{a-1})} - \frac{1}{4(y+\sqrt{a-1})} + \frac{1}{4 \sqrt{a-1}} \ln \left(\frac{y-\sqrt{a-1}}{y+\sqrt{a-1}}\right)$
Thus the integral is
$ \frac{1}{4(\sqrt{a}-\sqrt{a-1})} + \frac{1}{4(\sqrt{a}+\sqrt{a-1})} - \frac{1}{4 \sqrt{a-1}} \ln \left(\frac{\sqrt{a}-\sqrt{a-1}}{\sqrt{a}+\sqrt{a-1}}\right)$
which simplifies to
$ \frac{\sqrt{a}}{2} - \frac{1}{2 \sqrt{a-1}} \ln(\sqrt{a} - \sqrt{a-1})$
This is actually valid for all $a > 0$ except $a=1$, but for $a < 1$ you may want to use an alternative form to avoid complex numbers:
$ \frac{\sqrt{a}}{2} + \frac{1}{2 \sqrt{1-a}} \arctan\left(\sqrt{1/a-1}\right)$
For $a=1$ you can get the answer $1$ directly, or by taking the limit of either of these as $a \to 1$.