Question

∫5x+6/(x^2+x+1)

dx

Answer 1

分母的凑微分d(x²+x+1)=(2x+1)dx
所以将分子变形为
5x+6
=5*[(2x+1)-1]/2+6
=(5/2)(2x+1)+7/2
而x²+x+1=(x+1/2)²+3/4
所以∫ (5x+6)/(x²+x+1) dx
= ∫ [(5/2)(2x+1)+7/2]/(x²+x+1) dx
= (5/2)∫ (2x+1)/(x²+x+1) dx + (7/2)∫ 1/(x²+x+1) dx
= (5/2)∫ d(x²+x+1)/(x²+x+1) + (7/2)∫ d(x+1/2)/[(x+1/2)²+3/4]
= (5/2)ln( x²+x+1 ) + (7/2)*(2/√3)arctan[ (2x+1)/√3 ] + C
= (5/2)ln( x²+x+1 ) + (7/√3)arctan[ (2x+1)/√3 ] + C

后面那项用公式∫ dx/(x²+a²) = 1/a*arctan(x/a) + C