$$\int \frac{x^2+x+1}{3x^2-2x-5}$$
I can see where t start, maybe $$\int \frac{x^2+x+1}{3x^2-2x-5}=\int \frac{6x-2}{3x^2-2x-5}+\frac{x^2-5x+3}{3x^2-2x-5}=\ln|3x^2-2x-5|+\int \frac{x^2-5x+3}{3x^2-2x-5}$$
2026-03-31 13:48:17.1774964897
integrate $\int \frac{x^2+x+1}{3x^2-2x-5}$
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Hint:
Long division:
$$\frac{49}{24}\int\frac{dx}{3x-5}-\frac 1 8 \int \frac{dx}{x+1}+\frac 1 3 \int 1dx=\dots$$