- 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)
- 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)].