求有理函数的不定积分:∫x/x2+x+1 dx
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求有理函数的不定积分:∫x/x2+x+1 dx
∫x/(x^2+x+1) dx
= (1/2)∫ dln(x^2+x+1) -(1/2)∫1/(x^2+x+1) dx
= (1/2)ln(x^2+x+1) -(1/2)∫1/(x^2+x+1) dx
x^2+x+1= (x+1/2)^2+3/4
let
x+1/2 = (√3/2) tana
dx =(√3/2) (seca)^2 da
∫1/(x^2+x+1) dx
=∫(1/[(3/4)(seca)^2] ) (√3/2) (seca)^2 da
=(2√3/3)∫ da
=(2√3/3)a + C'
=(2√3/3) arctan[(2x+1)/√3]+ C'
∫x/(x^2+x+1) dx
= (1/2)ln(x^2+x+1) -(1/2)∫1/(x^2+x+1) dx
= (1/2)ln(x^2+x+1) -(√3/3) arctan[(2x+1)/√3]+ C
= (1/2)∫ dln(x^2+x+1) -(1/2)∫1/(x^2+x+1) dx
= (1/2)ln(x^2+x+1) -(1/2)∫1/(x^2+x+1) dx
x^2+x+1= (x+1/2)^2+3/4
let
x+1/2 = (√3/2) tana
dx =(√3/2) (seca)^2 da
∫1/(x^2+x+1) dx
=∫(1/[(3/4)(seca)^2] ) (√3/2) (seca)^2 da
=(2√3/3)∫ da
=(2√3/3)a + C'
=(2√3/3) arctan[(2x+1)/√3]+ C'
∫x/(x^2+x+1) dx
= (1/2)ln(x^2+x+1) -(1/2)∫1/(x^2+x+1) dx
= (1/2)ln(x^2+x+1) -(√3/3) arctan[(2x+1)/√3]+ C
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