Evaluate $$\int \frac{\arctan^6(3x)}{9x^2+1}dx\,.$$
My Attempt.
Let $u=\arctan(3x)$. Then $du = \frac{1}{9x^2+1}(3x)^\prime = \frac{3}{9x^2+1}$ and $$ \int \frac{\arctan^6(3x)}{9x^2+1}dx = \int \frac{u^6}{3}du = \frac{u^7}{21} + C = \frac{\arctan^7(3x)}{21} + C\,. $$ Is this correct?
Your solution is correct. If you want a sanity check, you can get the same result with integration by parts viz.$$u=\arctan^6(3x),\,v=\tfrac13\arctan(3x).$$