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[a,b] 表示下限a和上限b
∫ ln 1/(9-x^2) = - ∫ ln (9-x^2)
= - ∫ (ln (3-x) +ln (3+x) dx
= - ∫[0 ,3 ] (ln (3-x) dx - ∫[0,3] ln (3+x) dx
= ∫[0 ,3 ] (ln (3-x) d(3-x) - ∫[0,3] ln (3+x) d(3+x)
换元m = 3-x , n =3+x
= ∫[3 ,0 ] ln mdm - ∫[3,6] ln n dn
=- ∫[0,6 ] ln xdx
因为
∫lnxdx=xlnx-∫xd(lnx)
=xlnx-∫dx
=x(lnx-1)+C
所以
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x=0无意义
瑕积分
原式等于= x(lnx-1) | 从0到6
lnx 在 1 展开
lnx = ln(x-1+1)
ln[(x-1)+1]=(x-1)-1/2(x-1)^2+1/3(x-1)^3- ...+(-1)^(n-1)1/n(x-1)^n+o[(x-1)^n]
所以 lnx = x-1 +o(x-1)
所以
lim[x->0] x(lnx-1)
= lim[x->0] x(lnx-1) = x( x-1 +1)
=lim [x->0]x^2
=0
即原式等于=6ln6-6
[a,b] 表示下限a和上限b
∫ ln 1/(9-x^2) = - ∫ ln (9-x^2)
= - ∫ (ln (3-x) +ln (3+x) dx
= - ∫[0 ,3 ] (ln (3-x) dx - ∫[0,3] ln (3+x) dx
= ∫[0 ,3 ] (ln (3-x) d(3-x) - ∫[0,3] ln (3+x) d(3+x)
换元m = 3-x , n =3+x
= ∫[3 ,0 ] ln mdm - ∫[3,6] ln n dn
=- ∫[0,6 ] ln xdx
因为
∫lnxdx=xlnx-∫xd(lnx)
=xlnx-∫dx
=x(lnx-1)+C
所以
````````````````````````````````````````````````````````````
x=0无意义
瑕积分
原式等于= x(lnx-1) | 从0到6
lnx 在 1 展开
lnx = ln(x-1+1)
ln[(x-1)+1]=(x-1)-1/2(x-1)^2+1/3(x-1)^3- ...+(-1)^(n-1)1/n(x-1)^n+o[(x-1)^n]
所以 lnx = x-1 +o(x-1)
所以
lim[x->0] x(lnx-1)
= lim[x->0] x(lnx-1) = x( x-1 +1)
=lim [x->0]x^2
=0
即原式等于=6ln6-6
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