Given $ I_n = \int_0^1 \frac{(x^2 + x + 1)^n - x}{x^2 + 1} dx$ show that we have $I_{4n+1} \in \mathbb{Q}$, for any $n \in \mathbb{N}$. Consider the integral:
$$I_n = \int_0^1 \frac{(x^2+x+1)^n - x}{x^2 + 1} dx$$
I have to show 
$$I_{4n+1} \in \mathbb{Q}$$
for any $n \in \mathbb{N}$. I wasn't able to see any "direct" way of doing this, so I thought about induction. However, I couldn't find a way of expressing $I_{4n+1}$ in terms of $I_{4n-3}$ in order to use the induction step, so maybe induction is not the way either.
So how should I approach this?

Note that 
$$(x^2+x+1)^n=\big((x^2+1)+x\big)^n=\sum_{k=0}^n\binom{n}k(x^2+1)^kx^{n-k}\;,$$
so the integrand is
$$\frac{\sum_{k=0}^n\binom{n}k(x^2+1)^kx^{n-k}-x}{x^2+1}=\frac{x^n-x}{x^2+1}+\sum_{k=1}^n\binom{n}k(x^2+1)^{k-1}x^{n-k}\;.$$
The summation is a polynomial with integer coefficients, so its integral from $0$ to $1$ is rational, and we need only worry about the first term. When the exponent is $4n+1$ rather than $n$, it becomes
$$\begin{align*}
\frac{x(x^{4n}-1)}{x^2+1}&=\frac{x}{x^2+1}\big((x^4)^n-1^n\big)\\
&=\frac{x}{x^2+1}(x^4-1)\sum_{k=0}^{n-1}x^{4k}\\
&=\frac{x}{x^2+1}(x^2+1)(x^2-1)\sum_{k=0}^{n-1}x^{4k}\\
&=x(x^2-1)\sum_{k=0}^{n-1}x^{4k}\;,
\end{align*}$$
another polynomial with integer coefficients.
