We can expand the expression inside the integral $\left(x^2+1\right)^5$ using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer $n$. The formula is as follows: $\displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n$. The number of terms resulting from the expansion always equals $n + 1$. The coefficients $\left(\begin{matrix}n\\k\end{matrix}\right)$ are combinatorial numbers which correspond to the nth row of the Tartaglia triangle (or Pascal's triangle). In the formula, we can observe that the exponent of $a$ decreases, from $n$ to $0$, while the exponent of $b$ increases, from $0$ to $n$. If one of the binomial terms is negative, the positive and negative signs alternate.

Expand the integral $\int\left(x^{10}+5x^{8}+10x^{6}+10x^{4}+5x^2+1\right)dx$ into $6$ integrals using the sum rule for integrals, to then solve each integral separately

Gather the results of all integrals

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$