设F(x)是f(x)的一个原函数，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)
热心网友
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)].
热心网友
不行了，好多年没积分了，看着晕阿