设∫f’(t)dx=x(e^t+1)+C,则f(x)=?其中t=x^0.5,
1个回答

x=t^2

dx=2tdt

∫f’(t)dx=∫2tf’(t)dt=t^2(e^t+1)+C

设g'(t)=2tf'(t)

g(t)=t^2(e^t+1)+C

g'(t)=(t^2+2t)e^t+2t

f'(t)=(t/2+1)e^t+1

f(t)=∫f'(t)dt=∫[(t/2+1)e^t]dt+∫dt=∫(t/2+1)d(e^t)+t+C=(t/2+1)e^t-∫e^td(t/2+1)+t+C=(t+1)/2*e^t+t+C

f(x)=(x^0.5+1)/2*e^(x^0.5)+x^0.5+C

ps:偶只是高中生,可能解麻烦了,望见谅.