I dont think finding f(2) is possible....
f'(x)=x+f(x)--------->
f'''(x)=(x)''+f''(x)=f''(x)----->
f''(x)=c1*ex------->
f(x)=c1*ex+c2*x+c3
f(1)=c1*e+c2+c3=2-------------->
c3=2-c2-c1*e
f(2)=c1*e2+c2*2+c3=
c1*e2+c2+2-c1*e
edit: OH SHOOT I FORGOT SOMETHING
f'(x)=c2+c1*ex=x+c3+c2*x+c1*ex | subtract c1*ex ----->
c2=x+c3+c2*x. therefore, c2=-1(the only option. that means c3=c2=-1 and that means
f(1)=c1*e-2=2 therefore c1=4/e
that means f(2)=4*e-2-1=4e-3.