分类：教育/科学-升学入学-考研
问题：一道定积分题,寻求帮助设F(x)是f(x)的一个原函数，且当x& 
设F(x)是f(x)的一个原，且当x>0时，满足f(x)F(x)=ｘｅ^x/2(x+1)^2,F(x)<0,F(0)=-1求f(x)(x>0)
回答：f(x)F(x)=ｘｅ^x/2(x+1)^2=F'(x)F(x)==>
∫{0≤u≤x}F'(u)F(u)du=∫{0≤u≤x}uｅ^udu/[2(u+1)^2]=
==>F(x)^2/2-F(0)^2/2=
=∫{0≤u≤x}ｅ^udu/2(u+1)-∫{0≤u≤x}ｅ^udu/[2(u+1)^2]=
=∫{0≤u≤x}ｅ^udu/2(u+1)+∫{0≤u≤x}ｅ^ud{1/[2(u+1)]}=
=ｅ^x/[2(x+1)]-1/2+∫{0≤u≤x}ｅ^udu/2(u+1)-∫{0≤u≤x}ｅ^udu/2(u+1)=
=ｅ^x/[2(x+1)]-1/2==>
F(x)^2=ｅ^x/(x+1)==>F(x)=-ｅ^(x/2)/√(x+1)==>
f(x)=F'(x)=-[xｅ^(x/2)]/[2(x+1)^(3/2)].