∫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
驗算：
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
驗算正確.