菜鸟求助各位大神 高数

(3),做个代换：y=ux,则有x=1/u^2+1/u-1/(u-1),dx/du=-2/u^3-1/u^2+1/(u-1)^2
1/y^2=u^2(1-u)^2,所以有：
∫dx/y^2=∫(dx/du)[u^2(1-u)^2]du=∫[-2/u^3-1/u^2+1/(u-1)^2][u^2(1-u)^2]du
展开即可积分了。

∫dx/y^2=∫(dx/du)[u^2(1-u)^2]du=∫[-2/u^3-1/u^2+1/(u-1)^2][u^2(1-u)^2]du
=-2∫(1/u-2+u)du-∫(1-u)^2du+∫u^2du
=-2[lnu-2u+u^2/2]+(1-u)^3/3+u^3/3+C
=-2ln(y/x)+4y/x+y^2/2x^2+(1-y/x)^3/3+y^3/3x^3+C

4.
=|t|∫(0,1)|t-x|dx=|t|[∫(0,t)(t-x)dx+∫(t,1)(x-t)dx]
=|t|[t^2+(1-t)^2]/2=|t|(t^2-t+1/2)

令J=lim(n->∞)(a^n+c)/(b^n+d),假设a>0,b>0,c>0,d>0
(一）a<1:
b<1,J=(0+c)/(0+d)=c/d,
b=1,J=(0+c)/(1+d)=c/(1+d)
b>1,J=0
(二)a=1:
b<1,J=(1+c)/d
b=1,J=(1+c)/(1+d)
b>1,J=0
(三) a>1:
b<1,J=+∞
b=1,J=+∞
b>1:
b<a,J=+∞
b=a,J=1
b>a,J=0