Question

求大神解决不定积分

Answer 1

∫(3x+1)/(x^2+2x+17) dx
=(3/2)∫ (2x+2)/(x^2+2x+17) dx - 2∫ dx/(x^2+2x+17)
= (3/2)ln|x^2+2x+17| - 2∫ dx/(x^2+2x+17)

x^2+2x+17 = (x+1)^2 + 16
let
x+1= 4tany
dx = 4(secy)^2dy

∫ dx/(x^2+2x+17)
=(1/4)∫ dy
=(1/4)y+ C'
=(1/4)arctan[(x+1)/4] + C'

∫(3x+1)/(x^2+2x+17) dx
= (3/2)ln|x^2+2x+17| - 2∫ dx/(x^2+2x+17)
= (3/2)ln|x^2+2x+17| - (1/2)arctan[(x+1)/4] + C

Answer 2

原式=∫(3x+3-2)dx/(x^2+2x+17)
=3/2*∫(2x+2)dx/(x^2+2x+17)-2∫dx/(x^2+2x+17)
=3/2*∫d(x^2+2x+17)/(x^2+2x+17)-2∫dx/[(x+1)^2+16]
=3/2*ln(x^2+2x+17)-2*1/16∫dx/[(x/4+1/4)^2+1]
=3/2*ln(x^2+2x+17)-1/8*4∫d(x/4+1/4)/[(x/4+1/4)^2+1]
=3/2*ln(x^2+2x+17)-1/2*arctan(x/4+1/4)+C