设f(x)=ax^2+bx+cf(-1)=a-b+c=2f'(x)=2ax+bf'(0)=b=0∫0-1 f(x)dx=(a/3)x^3+(b/2)x^2+cx|0-1=a/3+b/2+c= -2所以 a=6 b=0 c=-4∫1-2 f(x)/x dx=∫1-2 (6x - 4/x) dx = 3x^2 - 4*lnx|1-2=(3*2*2-3*1*1)-4*(ln2-ln1)=9-...