∫1/(x^2+5x+6)dx

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/24 06:09:16

∫1/(x^2+5x+6)dx

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

求不定积分∫(1/x^2+6x+5)dx ∫1/(x^2+5x+6)dx ∫(x+1)/(x^2-2x+5)dx ∫[（x+1)/(x^(2)+2x+5]dx 求下列不定积分:∫(1+2x)/[x(x+1)]dx 和∫1/(X²-X-6)dx ∫1/(x^2+x+1)dx ∫X^2/1-x^2 dx. ∫dx/x^2(1-x^2) ∫dx/x^2(1+x^2) ∫x/（x^2+5）dx ∫（x^2+1/x^4)dx ∫(1-x)^2/x^3 dx