x<0
原式=0
0≤x≤1
原式=∫(-∞,0)f(t)dt+∫(0,x)f(t)dt
=∫(-∞,0)0dt+∫(0,x)tdt
=0+1/2t^2|(0,x)
=1/2 x^2
x>1
原式=∫(-∞,0)f(t)dt+∫(0,1)f(t)dt+∫(1,x)f(t)dt
=0+∫(0,1)tdt+∫(1,x)1/t√(t-1)dt
下解:∫(1,x)1/t√(t-1)dt
令√t-1=u
t=u^2 +1
dt=2udu
原式=∫(0,√(x-1)) 2u/[(u^2+1)u]du
=∫(0,√(x-1)) 2/(u^2+1)du
=2arctanu|(0,√(x-1))
=2arctan√(x-1)
所以
原式=1/2 +2arctan√(x-1)
追问对不起,是铅笔写的那个