<s>[INST] 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?
 [/INST] Sligthly differently:  $$I=\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
$$I=3\int_{0}^{1} \frac{x^3}{3x^2-3x+1}+\int_{0}^{1} 1 dx$$
to get $$ J=\int_{0}^{1} \frac{x^3}{3x^2-3x+1}dx~~~~(1)$$
Now use $\int_{0}^{a} f(x)= \int_{0}^{1} f(a-x) dx= 3J+1$
$$J=\int_{0}^{1} \frac{(1-z)^3}{3x^2-3x+1}dx~~~(2)$$
we get $$2J=\int_{0}^{1} 1 dx \implies J=\frac{1}{2} \implies I=\frac{5}{2}$$
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