1/(x²+2x+2)dx=1/[1+(x+1)²]dx=1/[1+(x+1)²]d(x+1) 令x+1=t 1/(1+t²）dt 的积分就是arctant 所以不定积分是arctan(x+1)+c x是0 到a 所以,值是arctan(a+1)-π/4=π/12 arctan(a+1)=π/3 a+1=tan(...