Find limit in use of integrals

We'll simplify with $y=nx$. Integration by parts gives $$\int y^5\arctan ydy=\frac{y^{6}}{6}\arctan y-\frac{1}{6}\int\frac{y^{6}}{1+y^{2}}dy\\=\frac{y^{6}}{6}\arctan y-\frac{1}{6}\int\left(y^{4}-y^{2}+1-\frac{1}{1+y^{2}}\right)dy\\=\frac{y^6+1}{6}\arctan y-\frac{1}{30}y^5+\frac{1}{18}y^3-\frac16 y+C.$$Hence $$\frac{1}{n^6}\int_{-n}^n y^5\arctan ydy=\frac{\frac{n^6}{3}\arctan n+o(n^6)}{n^6}\stackrel{n\to\infty}{\to}\frac{1}{3}\arctan\infty=\frac{\pi}{6}.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2\lim_{n\to\infty}\int_0^1 x^5\arctan nxdx$ (since the integrand is even), which by dominated convergence is $$\pi\int_0^1 x^5dx=\frac{\pi}{6}.$$

Hint: For the integral $$\int_{-1}^{1}x^5\arctan(nx)dx$$ we get $$\frac{15 \left(n^6+1\right) \tan ^{-1}(n)-3 n^5+5 n^3-15 n}{45 n^6}$$