Integral of 1/ sqrt(x^2+x+1)dx

integral of 1/ sqrt(x^2+x+1)dx

Start by completing the square:
∫ dx/√(x^2 + x + 1) = ∫ dx/√((x + 1/2)^2 + 3/4).

Now, let x + 1/2 = (√3/2) tan t,
dx = (√3/2) sec^2(t) dt.

So, we obtain
∫ (√3/2) sec^2(t) dt / ((√3/2) sec t)
= ∫ sec t dt
= ln |sec t + tan t| + C.

Finally, since tan t = (x + 1/2) / (√3/2), 'sohcahtoa' yields
ln |√(x^2 + x + 1)/(√3/2) + (x + 1/2)/(√3/2)| + C'
= ln |√(x^2 + x + 1) + (x + 1/2)| + C, where C = C' - ln(√3/2).

I hope this helps!