If $I=\int x^{27}\left(6x^2+5x+4\right){\left(x^2+x+1\right)}^{6}dx=f\left(x\right)+C$, then $f\left(x\right)$ is equal to

Solution:

We have

$$\begin{gathered} I=\int x^{27}\left(6x^2+5x+4\right){\left(x^2+x+1\right)}^{6}dx \\ \begin{array}{l}\Rightarrow I=\int \left(6x^5+5x^4+4x^3\right){\left(x^6+x^5+x^4\right)}^{6}dx\\ \Rightarrow I=\int {\left(x^6+x^5+x^4\right)}^{6}d\left(x^6+x^5+x^4\right)=\frac{1}{7}\left(x^6+x^5+x^4\right)+C\end{array} \end{gathered}$$

Hence, $f\left(x\right)=\frac{1}{7}{\left(x^6+x^5+x^4\right)}^7$.