Question

反常积分∫1/（x²+x-2）上限是正无穷大，下限是负无穷大 求解！

瑕点怎么处理

Answer 1

原式=∫1/(x+2)(x-1) dx
=1/3*∫[1/(x-1)-1/(x+2)] dx
=1/3*[ln|x-1|-ln|x+2|]
=1/3*ln|(x-1)/(x+2)]
=1/3*ln|(1-1/x)/(1+2/x)]

所以x趋于+∞和-∞
ln|(1-1/x)/(1+2/x)]极限都是ln(1/1)=0
所以原式=1/3*(0-0)=0

追问

瑕点1和2怎么处理

Answer 2

∫1/（x²+x-2）=∫1/[（x+2）(x-1)]dx=(1/3)[∫1/（x-1）dx-∫1/（x+2）dx==(1/3)ln|x-1|/|x+2|(上限是正无穷大，下限是负无穷大)
x趋向正无穷大时lim(1/3)ln|x-1|/|x+2|=lim(1/3)ln（x-1）/（x+2）=0
x趋向负无穷大时lim(1/3)ln|x-1|/|x+2|=lim(1/3)ln（-x+1）/（-x-2）=0
故∫1/（x²+x-2）=0

追问

瑕点1和2怎么处理

Answer 3

∫dx/(x^2+x-2)=∫dx/[(x+1/2)^2-9/4]=(2/3)∫d(2x/3 +1/3)/[(2x/3+1/3)^2-1]
2x/3+1/3=secu, 2dx/3=[sinu/cosu^2 ]du
原式=∫[sinu/cosu^2]du/[sinu/cosu]=∫du/cosu=∫dsinu/[1-(sinu)^2]=(1/2)[∫dsinu/(1-sinu)+∫dsinu/(1+sinu)]=(1/2)ln[(1+sinu)/(1-sinu)]+C
x→∞, u→0, x从-∞ →+∞ 时,u从0-→0+
∫[-∞,+∞]dx/(x^2+x-2)=∫[0-,0+][sinu/cosu^2]du/(sinu/cosu)=(1/2)(ln1-ln1)=0