设∫f'(x^3)dx=x^3+c,则f(x)等于

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设∫f''(x^3)dx=x^3+c,则f(x)等于设∫f''(x^3)dx=x^3+c,则f(x)等于设∫f''(x^3)dx=x^3+c,则f(x)等于∫f''(x³)dx=x³+Cf

设∫f'(x^3)dx=x^3+c,则f(x)等于
设∫f'(x^3)dx=x^3+c,则f(x)等于

设∫f'(x^3)dx=x^3+c,则f(x)等于
∫f'(x³) dx=x³+C
f'(x³)=3x²
令u=x³
x=u^(1/3)
f'(u)=3[u^(1/3)]2=3u^(2/3)
∴f'(x)=3x^(2/3)
f(x)=(9/5)x^(5/3)+C
验算:
f(x)=(9/5)x^(5/3)+C
f'(x)=9/5*5/3*x^(2/3)
=3x^(2/3)
f'(x³)=3(x³)^(2/3)=3x²
∫f'(x³) dx=3*x³/3+C
=x³+C
验算正确.

f(x)=(9/5)y^(5/3)+cy+k k为任意常数

1/4x^4+cx