Question

1÷(x^2+x+1)从正无穷到负无穷的无限积分

Answer 1

1÷(x^2+x+1)dx

=ʃdx/[(x+1/2)^2+3/4]

=ʃ4/3*dx/[(2x/√3+1/√3)^2+1]

=4/3*√3/2ʃd(2x/√3+1/√3)/[(2x/√3+1/√3)^2+1]

=2√3arctan(2x/√3+1/√3)+C

∴ʃ(-∞,+∞)1÷(x^2+x+1)dx

=2√3arctan(2x/√3)|(-∞,+∞)

=2√3π