∫x/(x^2+5x+6)dx

来源：学生作业帮编辑：神马作文网作业帮分类：综合作业时间：2024/11/10 23:39:25

∫x/(x^2+5x+6)dx

∫ x/(x+2)(x+3)dx
=∫ x ×[1/(x+2) -1/(x+3)] dx
=∫ x/(x+2)-x/(x+3) dx
=∫ 1-2/(x+2)-1+3/(x+3) dx
=∫ 3/(x+3)-2/(x+2) dx
=3ln|x+3|-2ln|x+2| +C
再问：请问=∫ 1-2/(x+2)-1+3/(x+3) dx这一步是什么意思？
再答： x/(x+2)
=[(x+2)-2]/(x+2)
=(x+2)/(x+2) -2/(x+2)
=1-2/(x+2) 能看懂吗

实际上用待定系数法
直接可把x/(x+2)(x+3) 化成3/(x+3) -2/(x+2)

∫[(x^2-x+6)/(x^3+3x)]dx ∫x/（x^2+5）dx 求不定积分∫(1/x^2+6x+5)dx ∫1/(x^2+5x+6)dx ∫(3x-2)/(x^2-2x+5)dx ∫(x+1)/(x^2-2x+5)dx ∫[（x+1)/(x^(2)+2x+5]dx ∫2x²+3x-5/x+3dx ∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=? ∫ [(x^3-2x^2+x+1)/(x^4+5x^2+4)]dx 不定积分! ∫dx/(根号[(x-a)(b-x)]) ∫xdx/根号(5+x-x^2) ∫csc x dx 求下列不定积分:∫(1+2x)/[x(x+1)]dx 和∫1/(X²-X-6)dx