Question

设f'(tanx+1)=(cosx)^2+(secx)^2,且f(1)=2,求f(x)

Answer 1

f '(tanx+1)=cos²x+sec²x=1/(1+tan²x)+1+tan²x

令t=tanx+1,则tanx=t-1

f '(t)=1/[1+(t-1)²]+1+(t-1)²

两端同时积分得

f(t)=∫1/[1+(t-1)²]dt +∫[1+(t-1)²]dt

=arctan(t-1)+t+(t-1)³/3+C

把f(1)=2代入得

0+1+0+C=2,解得C=1

故f(x)=arctan(x-1)+x+(x-1)³/3+1