Question

定积分计算问题~~急！

见图片，下界是负无穷，上界是正无穷，计算题

Answer 1

解：∵原式=∫(-∞，+∞)dx/[(x+1/2)²+3/4]²
∴设 x+1/2=√3/2tant
则dx=√3/2sec²tdt
且当x=-∞时，t=-π/2.当x=+∞时，t=π/2
故代入原式整理，得
原式=8√3/9∫(-π/2，π/2)cos²tdt
=8√3/9∫(-π/2，π/2)(1+cos(2t))/2dt
=4√3/9[t+sin(2t)/2]|(-π/2，π/2)
=4√3/9[π/2-(-π/2)+0-0]
=4√3π/9

Answer 2

1/3*(2*x+1)/(x^2+x+1)+4/9*3^(1/2)*atan(1/3*(2*x+1)*3^(1/2))|负无穷，正无穷
=4/9*pi*3^(1/2)