Question

高数问题求解

Answer 1

设t=x^3, x=t^(1/3)
dx=1/3*t^(-2/3)dt
∫f'(x^3)dx
=∫f'(t)*1/3*t^(-2/3)dt
=t+C
可以得出
1/3*f'(t)*t^(-2/3)=1
f'(t)=3t^(2/3)
f(t)=9/5*t^(5/3)+C
又f(1)=9/5
t=x^3=1,C=0
所以
f(t)=9/5*t^(5/3)
即
f(x)=9/5*x^(5/3)

Answer 2

∫f'(x³)dx=x³＋C
f'(x³)=3x²=3(x³)^(2/3)
f'(x)=3x^(2/3)
f(x)=9/5 x^(5/3) ＋C1
由于f(1)=9/5，所以C1=0
所以f(x)=9/5 x^(5/3)