x+1/x方+4x+5的dx.求积分

∫ (x+1)/(x^2+4x+5) dx
x^2+4x+5 = (x+2)^2 +1
let
x+2 = tany
dx = (secy)^2 dy
∫(x+1)/(x^2+4x+5) dx
= ∫(tany -1) dy
= -ln|cosy| - y + C
=-ln|1/√(x^2+4x+5)| - arctan(x+2) + C