if given the problem $$\int \frac{(27x^2)}{(4+9x^2)^\frac{3}{2}}$$ you would use a trig sub of $x = \frac{2}{3} \tan(\theta)$
after solving, I got an answer like this $$\frac{2}{3}(\ln|\frac{3x}{2} + \frac{3\sqrt{\frac{4}{9}+x^2}}{2}| - \frac{x}{\sqrt{\frac{4}{9}+x^2}})$$
I saw on wolfram alpha a similar answer using inverse hyperbolic sin but my course doesn't cover that. I feel as though im missing something is everything okay?
We have$$\int \frac{(27x^2)}{(4+9x^2)^\frac32}dx\\=\frac{27}8\int\frac{x^2}{(1+\frac94x^2)^\frac32}dx\\=\frac{27}8\int\frac{(\frac23tan\theta)^2(\frac23sec^2\theta)d\theta}{sec^3\theta}\\=\int\frac{(tan^2\theta)(sec^2\theta)d\theta}{sec^3\theta}\\=\int(sec\theta-cos\theta)d\theta\\=(ln|\frac{3x}{2} + \frac{3\sqrt{\frac{4}{9}+x^2}}{2}| - \frac{x}{\sqrt{\frac{4}{9}+x^2}})$$