已知f(X)在（-∞,+∞）连续,且f(x)=(根号x)/(1+x)+∫（下0上4）f（x）dx,求∫（下0上4）f（x

令M = ∫(0→4) f(x) dx,这是常数
f(x) = √x/(1 + x) + ∫(0→4) f(x) dx
==> f(x) = √x/(1 + x) + M
∫(0→4) f(x) = ∫(0→4) √x/(1 + x) dx + M∫(0→4) dx
M = ∫(0→4) 2x/(1 + x) d√x + 4M
- 3M = ∫(0→4) 2(x + 1 - 1)/(1 + x) d√x
- 3M = 2∫(0→4) [1 - 1/(1 + x)] d√x
- 3M = 2[√x - arctan√x]：(0→4)
- 3M = 4 - 2arctan(2)
M = (2/3)arctan(2) - 4/3
∴∫(0→4) f(x) dx = (2/3)arctan(2) - 4/3