高数，求不定积分的相关问题 10
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x+1= (1/宽伏2)(2x-2) +2
∫(x+1)/(x^2-2x+5) dx
=(1/2)∫(2x-1)/(x^2-2x+5) dx + 2∫dx/(x^2-2x+5)
=(1/2)ln|x^2-2x+5| + 2∫dx/(x^2-2x+5)
consider
x^2-2x+5 = (x-1)^2 +4
let
x-1= 2tany
dx= 2(secy)^2 dy
∫dx/(x^2-2x+5)
=(1/2)∫ dy
=(1/2)y + C'
=(1/2)arctan[(x-1)/2] + C'
∫(x+1)/(x^2-2x+5) dx
=(1/冲孝2)ln|x^2-2x+5| + 2∫dx/(x^2-2x+5)
=(1/散巧稿2)ln|x^2-2x+5| + arctan[(x-1)/2] + C
∫(x+1)/(x^2-2x+5) dx
=(1/2)∫(2x-1)/(x^2-2x+5) dx + 2∫dx/(x^2-2x+5)
=(1/2)ln|x^2-2x+5| + 2∫dx/(x^2-2x+5)
consider
x^2-2x+5 = (x-1)^2 +4
let
x-1= 2tany
dx= 2(secy)^2 dy
∫dx/(x^2-2x+5)
=(1/2)∫ dy
=(1/2)y + C'
=(1/2)arctan[(x-1)/2] + C'
∫(x+1)/(x^2-2x+5) dx
=(1/冲孝2)ln|x^2-2x+5| + 2∫dx/(x^2-2x+5)
=(1/散巧稿2)ln|x^2-2x+5| + arctan[(x-1)/2] + C
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