integration

Question

∫  4x²+3/(x²+x+1)² dx

Answer 1

(4x^2+3)/(x^2+x+1) =

= (4x^2+4-1)/(x^2+x+1)

= (4x^2+4)/(x^2+x+1)-1/(x^2+x+1)

= [4(x^2+1)/(x^2+x+1)]-[1/(x^2+x+1]

= 4 [(x²+1) /(x²+x+1)] - [1 /(x²+x+1)]

= 4 {[(x²+x+1)-x] /(x²+x+1)} - [1 /(x²+x+1)]

= 4 [(x²+x+1) /(x²+x+1)] - [4x/(x²+x+1)] - [1 /(x²+x+1)]

= 4 - [4x/(x²+x+1)] - [1/(x²+x+1)]

= 4 - [(4x+1) /(x²+x+1)]

Now the integral is[(4x²+3) /(x²+x+1)]dx = integral  {4 -[(4x+1) /(x²+x+1)]} dx

∫ (4x² + 3) /(x² + x + 1) = 4x - 2 ln (x² + x + 1) - (2/√3) arctan[(2/√3)√x + (1/√3)] +
(2/√3) arctan[(2/√3)√x - (1/√3)]} + C