对的
∫0.1dr/(1+0.1r)dr
=∫d0.1r/(1+0.1r)dr
=∫d(1+0.1r)/(1+0.1r)dr
=ln(1+0.1r)(t到0)
=ln(1+0.1t)-ln1
=ln(1+0.1t)
所以a(t)=e^[ln(1+0.1t)]=1+0.1t
对的
∫0.1dr/(1+0.1r)dr
=∫d0.1r/(1+0.1r)dr
=∫d(1+0.1r)/(1+0.1r)dr
=ln(1+0.1r)(t到0)
=ln(1+0.1t)-ln1
=ln(1+0.1t)
所以a(t)=e^[ln(1+0.1t)]=1+0.1t