Here is a solution using a substitution, and instead of polynomial long division we end up using a reduction when integrating tangent functions. I don't claim it is easier. It is just (slightly) different.
By the substitution $u=\arctan x$ the integral transforms into
Next, since $D\tan u=1+\tan^2u$ we find that for $n\geq 2$ (here, we should
interpret $\tan^0u$ as being equal to $1$)
Applying this reduction several times on
we end up finding that the integral $(*)$ equals
Here, it could be funny to note that while reducing the odd powers of $\tan u$, we do not get any $\int\tan u\,du$, a term which would have given us logarithms in the answer.