CarieVinne -
见图
二分好久没看 -
这里有递推公式的，方法特殊：
∫1/(x^2+x+1)dx (先用分部积分）
=x/(x^2+x+1)-∫x(-2x-1)/(x^2+x+1)^2 dx
=x/(x^2+x+1)-∫x(-2x-1)/(x^2+x+1)^2 dx
=x/(x^2+x+1)+2∫1/(x^2+x+1)dx-(1/2)∫(2x+1)/(x^2+x+1)^2dx-(3/2)∫1/(x^2+x+1)^2dx
所以：∫1/(x^2+x+1)^2dx
=(2/3)x/(x^2+x+1)+(2/3)∫1/(x^2+x+1)dx+(1/3)/(x^2+x+1)
=(1/3)(2x+1)/(x^2+x+1)+(2/3)∫1/((x+1/2)^2+3/4)dx
=(1/3)(2x+1)/(x^2+x+1)+(4/(3√3))arctan(2x+1)/√3+C
你这是干啥嘛 -
原式=∫ dx/[ (x+1/2)^2+3/4]^2
=∫ d(x+1/2)/{(3/4)[ 2(x+1/2)/√3]^2+1}^2
=(4/3)(√3/2)∫ d[2(x+1/2)/ √3]/{[ 2(x+1/2)/√3]^2+1}^2
设2(x+1/2)/√3为u,
原式=(2/√3)∫ du/(1+u^2)^2
设u=tant,du=(sect)^2dt,
t=arctanu,
cost=1/√(1+u^2),
sint=u/√(1+u^2),
原式=(2/√3)∫ (sect)^2dt/(sect)^4
=(2/√3)∫ (cos)^2dt
=(1/√3)∫ (1+cos2t)dt
=√3t/3+(√3/6)sin2t+C
=(√3/3)arctanu+(√3/3)*[u/√(1+u^2)][1/√(1+u^2)]+C
=(√3/3)arctan[(2x+1)/√3]+[(√3/3)(2x+1)/√3]/[1+(2x+1)^2/3]+C
=(√3/3)arctan[(2x+1)/√3]+(3/4)(2x+1)/(x^2+x+1)+C.
敬岭 -
∫1/(x^2+x+1)^2*dx=∫1/((x+1/2)^2+3/4)^2*dx
=16/9*∫1/(4/3*(x+1/2)^2+1)^2*dx
=16/9*∫1/((√3/3*(2x+1))^2+1)^2*dx
令√3/3*(2x+1)=tany, 则2√3/3*dx=secy^2*dy, dx=√3/2*secy^2*dy,
原式=16/9*∫1/(1+tany^2)^2*√3/2*secy^2*dy=8√3/9*∫1/( secy^2)^2*secy^2*dy
=8√3/9*∫1/( secy^2)*dy
=8√3/9*∫cosy^2*dy
=8√3/9*∫1/2*(cos2y+1)*dy
=4√3/9*∫(cos2y+1)*dy
=4√3/9*(1/2*sin2y+y) +常数
=2√3/9*sin2y+4√3/9*y) +常数
= 2√3/9*sin[2*arctan(√3/3*(2x+1))]+ 4√3/9* arctan(√3/3*(2x+1)) +常数