求定积分 ∫(上1下0)1/(x^2-2x-3)dx

来源:学生作业帮助网 编辑:六六作业网 时间:2024/05/15 12:23:39
求定积分∫(上1下0)1/(x^2-2x-3)dx求定积分∫(上1下0)1/(x^2-2x-3)dx求定积分∫(上1下0)1/(x^2-2x-3)dx∫[0,1]1/(x^2-2x-3)dx=1/5∫

求定积分 ∫(上1下0)1/(x^2-2x-3)dx
求定积分 ∫(上1下0)1/(x^2-2x-3)dx

求定积分 ∫(上1下0)1/(x^2-2x-3)dx
∫[0,1] 1/(x^2-2x-3)dx
=1/5∫[0,1] [1/(x-3)-1/(x+2)]dx
=1/5[ln|x-3|-ln|x+2|] [0,1]
=1/5(ln2-ln3-ln2+ln2)
=0

∫(上1下0)1/(x^2-2x-3)
= ∫(上1下0)1/(x+1)(x-3)
= ∫(上1下0)1/4[1/(x-3)-1/(x+1)]
=1/4 ∫(上1下0)[1/(x-3)-1/(x+1)]
=1/4 ∫(上1下0)1/(x-3)-1/4 ∫(上1下0)1/(x+1)
=-1/4ln3