Integration practice of $\int \frac{\sqrt{25-y^2}}{y}dy$ I need to solve $\int \frac{\sqrt{25-y^2}}{y}dy$.
I originally thought IBP, but that led to a very large and confusing algebra problem. 
Then I started to look at the $\sqrt{25-y^2}$ and started to consider doing a $u$-substituion, and letting $u=\sin\theta$ and $du=\cos\theta d\theta$. Once I started playing a bit, the intuition was lost.
Any and all help is appreciated.

Trigonometric substitutions are not necessary.  Let $u^2 = 25 - y^2$, so that $2u \, du = -2y \, dy$.  Then $$\begin{align*} \int \frac{\sqrt{25-y^2}}{y} \, dy &= - \int \frac{-y \sqrt{25-y^2}}{y^2} \, dy \\ &= - \int \frac{u \sqrt{u^2}}{25-u^2} \, du \\ &= \int \frac{u^2 - 25 + 25}{u^2 - 25} \, du \\ &= u + 25 \int \frac{du}{u^2 - 25} \\ &= u + \frac{5}{2} \int \left(\frac{1}{u-5} - \frac{1}{u+5} \,\right) du\end{align*}$$ and the rest is obvious.
