求定积分 ∫(上1下0)1/(x^2-2x-3)dx
网友回答
∫[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======以下答案可供参考======
供参考答案1:
∫(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