积分∫[(x+1)/(x²+x+1)]dx如何解,

积分∫[(x+1)/(x²+x+1)]dx如何解,
积分∫[(x+1)/(x²+x+1)]dx如何解,

积分∫[(x+1)/(x²+x+1)]dx如何解,

=1/2∫(2x+2)/(x^2+x+1)dx
=1/2∫(2x+1+1)/(x^2+x+1)dx
=1/2(∫(2x+1)/(x^2+x+1)dx+∫1/(x^2+x+1)dx)
=1/2(∫1/(x^2+x+1)d(x^2+x+1)+∫1/(x^2+x+1)dx)
=1/2(∫1/(x^2+x+1)d(x^2+x+1)+∫1/(x+1/2)^2+3/4)dx

全部展开

=1/2∫(2x+2)/(x^2+x+1)dx
=1/2∫(2x+1+1)/(x^2+x+1)dx
=1/2(∫(2x+1)/(x^2+x+1)dx+∫1/(x^2+x+1)dx)
=1/2(∫1/(x^2+x+1)d(x^2+x+1)+∫1/(x^2+x+1)dx)
=1/2(∫1/(x^2+x+1)d(x^2+x+1)+∫1/(x+1/2)^2+3/4)dx
=1/2(∫1/(x^2+x+1)d(x^2+x+1)+∫1/(x+1/2)^2+3/4)d(x+1/2))
=1/2(ln(x^2+x+1)+2/√3arctan((x+1/2)/(√3/2))+C
=1/2(ln(x^2+x+1)+2/√3arctan((2x+1)/√3)+C

收起

1/2∫(2x+2)/(x^2+x+1)dx
=1/2∫1/(x^2+x+1)d(x^2+x+1)
=1/2ln(x^2+x+1)+C