答：f(x) = 2sinx + cosx
f(x) = 1 + 2x + ∫(0~x) tf(t) dt - x∫(0~x) f(t) dt ...(1)
f'(x) = 2 + xf(x) - [∫(0~x) f(t) dt + xf(x)]
f'(x) = 2 - ∫(0~x) f(t) dt
f''(x) = -f(x)
f''(x) + f(x) = 0 ...(2)
特征方程：r² + 1 = 0 => r = ±i
f(x) = Asinx + Bcosx，A、B为任意常数
由(1)：f(0) = 1
=> f(0) = Asin(0) + Bcos(0) = B
=> B = 1
f(x) = Asinx + cosx，代入(1)：
Asinx + cosx = 1 + 2x + ∫(0~x) (t - x)(Asint + cost) dt
Asinx + cosx = 1 + 2x + Asinx + cosx - Ax - 1
=> A = 2
所以f(x) = 2sinx + cosx
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