Question

求不定积分： ∫(x+3)/(x^2-5x+6)dx=

Answer 1

∫(x+3)/(x²-5x+6)dx
= ∫(x+3)[(x-2)-(x-3)]/[(x-3)(x-2)]dx
= ∫(x+3)[1/(x-3)-1/(x-2)]dx
= ∫[6/(x-3)-5/(x-2)]dx
= 6ln(x-3)-5ln(x-2)+C

Answer 2

有理函数的积分,待定系数法
∫(2x-1)/(x-2)(x-3)
dx=∫a/(x-2)+b/(x-3)
dx
所以a(x-3)+b(x-2)=2x-1
a+b=2
-3a-2b=-1解得a=-3,b=5
∫(2x-1)/(x-2)(x-3)
dx=∫a/(x-2)+b/(x-3)
dx
=-3∫dx/(x-2)+5∫dx/(x-3)
=-3ln|x-2|+5ln|x-3|+c