∫(1,x)tf(t)dt=xf(x)+x^2,当x=1时,0=1*f(1)+1^2=f(1)+1,f(1)=-1,两边对x求导数xf(x)=f(x)+xf'(x)+2x,初值条件为f(1)=-1,解常微分方程可得,（MATLAB）y=dsolve('x*y=y+x*Dy+2*x','y(1)=-1')y =2/(x-1)*x-exp((x-1)*t...