Evaluate $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$ This integral
$$\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
is fabricated based on an interesting point which I may post later.
The question is: How would you do it?

\begin{gather*}
I=\int ^{1}_{0}\left(\frac{3x^{3}}{3x( x-1) +1} +1\right) dx=\int ^{1}_{0}\frac{3x^{3} dx}{3x( x-1) +1} +1\\
Let\ I_{1} =\int ^{1}_{0}\frac{x^{3} dx}{3x( x-1) +1}\\
I_{1} =\int ^{1}_{0}\frac{( 1-x)^{3} dx}{3( 1-x)( 1-x-1) +1} =\int ^{1}_{0}\frac{( 1-x)^{3} dx}{3x( x-1) +1}\\
because\int ^{a}_{0} f( x) dx=\int ^{a}_{0} f( a-x) dx\\
Adding\ both,\\
2I_{1} =\int ^{1}_{0}\frac{x^{3} +( 1-x)^{3} dx}{3x( x-1) +1}\\
Apply\ a^{3} +b^{3} =( a+b)\left( a^{2} +b^{2} -ab\right)\\
2I_{1} =\int ^{1}_{0}\frac{3x( x-1) +1dx}{3x( x-1) +1} =1\\
I_{1} =\frac{1}{2}\\
I=3I_{1} +1=\frac{3}{2} +1=\frac{5}{2}\\
\\
\end{gather*}which I hope is the correct answer:)
