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intergral 9/1+9x^2
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\[\Large \int\limits \frac{9}{1+9x^2}dx\]Looks very similar to arctan yes?\[\Large \int\limits \frac{1}{1+x^2}dx \quad=\quad \arctan x\]
Umm have you learned how to do trig subs?
Oh I guess we don't need a trig sub here :) lol. Let's just make a u-sub.
\[\Large \int\limits \frac{9}{1+9x^2}dx\quad=\quad 3\int\limits\frac{1}{1+(3x)^2}(3dx)\] \[\Large u=3x \qquad\to\qquad du=3dx\]
\[\Large = \quad 3\int\limits \frac{1}{1+(u)^2}(du)\]
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See how we can relate it back to the arctan integral? :o
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