1/(x^2+4x+9)
=1/[(x+2)^2+5]
原式=1/√5*arctan(x/√5)|(-∞,+∞)
=1/√5*[π/2-(-π/2)]
=π/√5 1/(x^2+4x+9)
=1/[(x+2)^2+5]
原式=1/√5*arctan(x/√5)|(-∞,+∞)
=1/√5*[π/2-(-π/2)]
=π/√5