Let y=x+½, y2=(x+½)2, dy/dx=1, so dy=dx;
x2+x+1=x2+x+¼+¾=(x+½)2+¾=y2+¾.
4x2+3=4(y-½)2+3=4(y2-y+¼)+3=4y2-4y+4 or 4(y2-y+1).
Let L=∫(4x2+3)/(x2+x+1)dx=4∫[(y2-y+1)/(y2+¾)2]dy.
y2-y+1=(y2+¾)-y+¼, so the integral can be split:
4∫dy/(y2+¾)-4∫[y/(y2+¾)2]dy+∫dy/(y2+¾)2.
Let I=4∫dy/(y2+¾); J=4∫ydy/(y2+¾)2; K=∫dy/(y2+¾)2.
Let y=½√3tanθ, y2=¾tan2θ, dy=½√3sec2θdθ;
y2+¾=¾tan2θ+¾=¾sec2θ; (y2+¾)2=(9/16)sec4θ.
I=4∫½√3sec2θdθ/(¾sec2θ)=(4/3)(2√3)θ+C1=(8√3/3)arctan(⅔y√3)+C1, where C1 is a constant.
Let z=y2+¾, dz=2ydy, J=2∫z-2dz, J=-2/z+C2=-2/(y2+¾)+C2, where C2 is another constant.
K=∫½√3sec2θdθ/(9/16)sec4θ,
K=(½√3)(16/9)∫cos2θdθ,
K=(4√3/9)∫(cos(2θ)+1)dθ.
K=(4√3/9)(½sin(2θ)+θ)+C3=(4√3/9)(sinθcosθ+θ)+C3, where C3 is a constant.
tanθ=⅔y√3=2y/√3, sinθ=⅔y√3/√(4y2/3+1), cosθ=1/√(4y2/3+1);
sinθcosθ=[⅔y√3/√(4y2/3+1)][1/√(4y2/3+1)],
sinθcosθ=⅔y√3/(4y2/3+1)=2y√3/(4y2+3);
θ=arctan(⅔y√3);
K=(4√3/9)(2y√3/(4y2+3)+arctan(⅔y√3))+C3,
K=8y/(12y2+9)+(4√3/9)arctan(⅔y√3)+C3,
Let constant C=C1-C2+C3, then:
L=I-J+K=(8√3/3)arctan(⅔y√3)+2/(y2+¾)+8y/(12y2+9)+(4√3/9)arctan(⅔y√3)+C;
8√3/3+4√3/9=28√3/9;
2/(y2+¾)+8y/(12y2+9)=2/(y2+¾)+⅔y/(y2+¾)=(2+⅔y)/(y2+¾)=(8/3)(3+y)/(4y2+3);
L=(28√3/9)arctan(⅔y√3)+(8/3)(3+y)/(4y2+3).
Finally putting y=x+½:
4y2+3=4(x+½)2+3=4(x2+x+1);
(8/3)(3+y)/(4y2+3)=⅓(2x+7)/(x2+x+1);
L=(28√3/9)arctan((2x+1)√3/3)+⅓(2x+7)/(x2+x+1).