Question

求解线性方程组

x/x²+x+1的积分

Answer 1

∫ x/(x^2+x+1) dx

=(1/2)∫ (2x+1)/(x^2+x+1) dx -(1/2)∫ dx/(x^2+x+1)
=(1/2)ln|x^2+x+1| -(1/2)∫ dx/(x^2+x+1)
=(1/2)ln|x^2+x+1| -(√3/3)arctan[(2x+1)/√3]+ C

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x^2+x+1 =(x+ 1/2)^2 + 3/4
let
x+1/2 =(√3/2)tanu
dx=(√3/2)(secu)^2 du
∫ dx/(x^2+x+1)
=∫ (√3/2)(secu)^2 du/[(3/4)(secu)^2 ]
=(2√3/3)u+ C
=(2√3/3)arctan[(2x+1)/√3]+ C