∫123x2+2x3+2x dx=\displaystyle\int_1^2 \frac{3x^2 + 2}{x^3 + 2x} \, dx = ∫12x3+2x3x2+2dx=
ln(3x2+2)−ln(x3+2x)\ln(3x^2 + 2) - \ln(x^3 + 2x)ln(3x2+2)−ln(x3+2x)
ln3\ln 3ln3
ln4\ln 4ln4
123\frac{12}{3}312
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