第1个回答 2019-08-09
x^2+2x+3 = (x+1)^2 +2
let
x+1 =√2 tanu
dx =√2 (secu)^2 du
∫(x-1)/(x^2+2x+3) dx
=(1/2)∫(2x+2)/(x^2+2x+3) dx - 2∫dx/(x^2+2x+3)
=(1/2)ln|x^2+2x+3| -2∫dx/(x^2+2x+3)
=(1/2)ln|x^2+2x+3| -2∫√2 (secu)^2 du/[ 2(secu)^2 ]
=(1/2)ln|x^2+2x+3| -√2u + C
=(1/2)ln|x^2+2x+3| -√2arctan[(x+1)/√2] + C