<s>[INST] A definite integral involving a parameter Does this integral seem to have a nice closed form (at least for a subset of values of $a > 0$)
$$
\int_{\sqrt{a}}^\infty \frac{y^2}{(y^2-a+1)^2} dy
$$
Using a symbolic math software and for $a$ being some small integers, I end up getting different answers in terms of $\sinh^{-1}$, $\log$, $\tanh^{-1}$ and $\coth^{-1}$ depending on $a$.
 [/INST] 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 $a \to 1$.
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