求下列不定积分:∫(1+2x)/[x(x+1)]dx 和∫1/(X²-X-6)dx

令(1 + 2x)/[x(x + 1)] = A/x + B/(x + 1)
令x = 0,A = (1 + 0)/(0 + 1) = 1
令x = - 1,B = (1 - 2)/(- 1) = 1
∴(1 + 2x)/[x(x + 1)] = 1/x + 1/(x + 1)
∫ (1 + 2x)/[x(x + 1)] dx
= ∫ 1/x dx + ∫ 1/(1 + x) dx
= ln| x | + ln| 1 + x | + C
= ln| x(1 + x) | + C
∫ 1/(x² - x - 6) dx
= ∫ 1/[(x + 2)(x - 3)] dx
= ∫ [(x + 2) - (x - 3)]/[(x + 2)(x - 3)] dx
= (1/5)∫ [1/(x - 3) - 1/(x + 2)] dx
= (1/5)(ln| x - 3 | - ln| x + 2 |) + C
= (1/5)ln| (x - 3)/(x + 2) | + C