If you apply long division, you get \(\displaystyle 4-\frac{36}{x^{2}+9}\)
\(\displaystyle \int{4dx}-36\int\frac{1}{x^{2}+9}dx\)
Notice the one on the right?. Looks a whole lot like a familiar integral. Perhaps an arctan.
Let \(\displaystyle x=3tan(u), \;\ u=tan^{-1}(\frac{x}{3}), \;\ du=\frac{3}{x^{2}+9}dx, \;\ \frac{du}{3}=\frac{1}{x^{2}+9}dx\)
Make the subs
