Respuesta :

∫㏑(x² - x + 2) dx = ∫㏑(x² - 2x + x + 2) dx
∫㏑(x² - x + 2) dx = ∫㏑[x(x) - x(2) + 1(x) - 1(2)] dx
∫㏑(x² - x + 2) dx = ∫㏑[x(x - 2) + 1(x - 2)] dx
∫㏑(x² - x + 2) dx = ∫㏑[(x + 1)(x - 2)] dx
∫㏑(x² - x + 2) dx = ∫㏑(x + 1) + ㏑(x - 2) dx
∫㏑(x² - x + 2) dx = ∫㏑(x + 1) dx + ∫㏑(x - 2) dx
∫㏑(x² - x + 2) dx = [x㏑(x + 1) - (x - ㏑(|x + 1|)) + C] + [x㏑(x - 2) - (x + ㏑(|x - 2|)) + C]
∫㏑(x² - x + 2) dx = [x ln(x + 1) + x㏑(x - 2)] + [(-x +㏑(|x + 1|)) + (-x - ln(|x - 2))] + {C + C}
∫㏑(x² - x + 2) dx = x㏑(x² - x + 2) + [-2x - (㏑(|x + 1|))/㏑(|x - 2|))] + 2C
∫㏑(x² - x + 2) dx = x㏑(x² - x + 2) - 2x - (㏑(|x + 1|))/㏑(|x - 2|)) + 2C