Question

设f'(x^2+2)=3x^2+1+5/(2+x^2)求f(x)

属于不定积分范畴的
答案是f(x)=1.5x^2-5x+c-5/x

Answer 1

令x^2+2=t, f'（t）=3t-5+5/t, f(t)=1.5t^2-5t+5lnt +C,f(x)=1.5x^2-5x+5lnx +C

Answer 2

f'(x²+2) = 3x²+1+5/(x²+2) = 3(x²+2-2)+1+5/(x²+2) = 3(x²+2)-5+5/(x²+2)
f'(x) = 3x-5+5/x
f(x) = ∫ (3x-5+5/x) dx = 3*x²/2 - 5x + 5ln|x| + C
= 1.5x² - 5x + 5ln|x| + C

Answer 3

∵f'(x^2+2)=3x^2+1+5/(2+x^2)=3(x^2+2)+5/(2+x^2)-5
∴令t=x^2+2
则f'(t)=3t+5/t-5
∴f(t)=1.5t^2+5lnt-5t+c
∴f(x)=1.5x^2+5lnx-5x+c

Answer 4

设x^2＋2＝t
得f“（t）＝3t＋（5／t）－5
3t＝（3／2）＊t逆导为1.5t^2
－5为－5t
5／t为5Int