Question

integrate dx/9+4x^2

Answer 1

start with
\[\int\frac{dx}{9+(2x)^2}\]

Answer 2

the gimmick is to make it look like
\[\frac{1}{1+u^2}\] so that
\[\int\frac{du}{1+u^2}=\tan^{-1}(u)\]

Answer 3

factor a 9 out of the denominator to get
\[\int\frac{dx}{9(1+(\frac{2}{3}x)^2)}\]
\[=\frac{1}{9}\int \frac{dx}{1+\left(\frac{2}{3}x\right)^2}\]

Answer 4

then use a u sub \(u=\frac{2}{3}x, du=\frac{2}{3} dx,\frac{3}{2}du = dx\) to get
\[\frac{1}{6}\int \frac{du}{1+u^2}=\frac{1}{6}\tan^{-1}(u)\]