Correct answer: (c) \(\frac{4}{3\sqrt{3}} tan^{-1}(\frac{2x+1}{\sqrt{3}}) + \frac{1}{3}(\frac{2x+1}{x^2+x+1})+c\)

Explanation:

I = \(\int \frac{1}{(x^2+x+1)^2}dx\) = \(\int \frac{1}{[(x + \frac{1}{2})^2 + \frac{2}{4}]^2}dx\)

Let x + \(\frac{1}{2}\) = a ⇒ da = dx

= \(\int \frac{1}{(a^2 + \frac{3}{4})^2}da\)

Let a = \(\frac{\sqrt 3}{2}\)tan θ ⇒ da = \(\frac{\sqrt 3}{2}sec^2 \theta \ d \theta\)

\((a^2 + \frac{3}{4})^2\) = \((\frac{3}{4}tan^2 \theta + \frac{3}{4})^2\)

⇒ \((a^2 + \frac{3}{4})^2\) = \(\frac{9 \ sec^4 \theta}{16}\)

= \(\frac{\sqrt 3}{2}\)\(\int \frac{16 \ d \theta}{9sec^2 \theta}\) = \(\frac{8}{3 \sqrt 3}\)\(\int cos^2 \theta d \theta\)

= \(\frac{8}{3 \sqrt 3}\) \(\int (\frac{1}{2}cos2 \theta + \frac{1}{2})d \theta\) = \(\frac{4}{3 \sqrt 3}(\frac{sin 2 \theta}{2} + \theta)\)

= \(\frac{4}{9}\)\(\Big[\frac{6a}{4a^2+3} + \sqrt 3 tan^{-1}(\frac{2a}{\sqrt 3}) \Big]\)

= \(\frac{4 \sqrt 3(x^2 + x + 1)tan^{-1} (\frac{2x+1}{\sqrt3})+6x+3}{9(x^2+x+1)}\)

= \(\frac{1}{9}\)\(\Big[\frac{6x+3}{x^2+x+1} + 4 \sqrt3 tan^{-1} (\frac{2x+1}{\sqrt3})\Big] \)

= \(\frac{4}{3\sqrt3}tan^{-1}\)\((\frac{2x+1}{\sqrt3})\) + \(\frac{1}{3}\)\(\Big(\frac{2x+1}{x^2+x+1}\Big)\)