解：∵x-∫<1,y+x>e^(-t²)dt=0 ==>1-(y'+1)e^(-(y+x)²)=0 (等式两端求导) ==>y'+1=e^(y+x)² ==>y'=e^(y+x)²-1 ∴dy/dx=y'=e^(y+x)²-1。
