Question

计算无穷限反常积分的解?上限为（无穷大）,下限为（无穷小）.∫[1/（x^2+x+1）]dx=?特别是定积分是怎么划开的

Answer 1

∫[1/（x^2+x+1）]dx=∫[1/（(x+1/2)^2+3/4）]dx=∫[1/（(x+1/2)^2+3/4）]dx

=4/3*∫[1/((2/√3*(x+1/2))^2+1）]dx

=2/√3*∫[1/((2/√3*(x+1/2))^2+1）]d(2/√3*x)

=2/√3*∫[1/((2/√3*(x+1/2))^2+1）]d(2/√3*(x+1/2))

=2/√3*arctan(2/√3*(x+1/2))

当x->+∞时,arctan(2/√3*(x+1/2))—>π/2

当x->-∞时,arctan(2/√3*(x+1/2))—>-π/2

所以,原定积分=2/√3（π/2+π/2）=2/√3*π=2√3*π/3