高数不定积分 f'(tanx)=sec^2x,且f(0)=1,求f(x)
问题描述:
高数不定积分 f'(tanx)=sec^2x,且f(0)=1,求f(x)
答案是x+x^3/3+C,怎么算的
答
ƒ'(tanx) = sec²x = 1 + tan²x
ƒ'(x) = 1 + x²
ƒ(x) = ∫ (1 + x²) dx = x + x³/3 + C
ƒ(0) = 1 ==> 0 + 0/3 + C = 1 ==> C = 1
∴ƒ(x) = x + x³/3 + 1