Indefinite Integral Inverse Trigonometric Function $$\int \frac{2}{9w^2+25}dw$$
I already know this will be equal to $\frac{1}{a} \arctan(x/a)$,
but I don't know how to factor out the $9$. I only know how to take out the $2$.

$$
\begin{align}
& \phantom{={}} 2\int \frac{dw}{9w^2+25} = 2\int\frac{dw}{25(\frac{9}{25} w^2 + 1)} = \frac{2}{25}\int\frac{dw}{\left(\frac35 w\right)^2 + 1} \\[12pt]
& = \frac{2}{3\cdot5} \int \frac{\frac35\, dw}{\left(\frac35 w\right)^2 + 1} = \frac{2}{15} \int \frac{du}{u^2+1}\text{ etc.}
\end{align}
$$
