<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] $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx=\int_{0}^{1} \frac{(x+2)(3(x^2-x)+1)+2x-1}{3(x^2-x)+1}dx =\int_{0}^{1} (x+2+\frac{2x-1}{3(x^2-x)+1})dx=\int_{0}^{1} (x+2+\frac{(x^2-x+\frac{1}{3})'}{3(x^2-x+\frac{1}{3})})dx=[\frac{1}{2}x^2+2x+\ln(x^2-x+\frac{1}{3})]_0^{1}=\frac{5}{2}$
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