Where was my mistake (integration by trig-substitution problem)? I am attempting to solve the problem
$$\int \frac{dx}{x^2+x+1}$$
First, I complete the square, then factor out a $\frac {3}{4}$:
$$\int \frac{dx}{\frac{3}{4}(\frac{4}{3}(x+\frac{1}{2})^2+1)}$$
Let $u = \sqrt{\frac{4}{3}}(x+\frac{1}{2})$
$$\frac{du}{dx} = \frac{2}{\sqrt{3}}$$
$$dx = \frac{\sqrt{3}}{2} du$$
Thus, we now have the integral:
$$\frac{4}{3} \frac{\sqrt{3}}{2} \int \frac{du}{u^2+1}$$
Let $u = \tan \theta$
$$du = \sec^2\theta \ d\theta$$
What follows is obvious now, and the solution should be:
$$\frac{4}{3} \frac{\sqrt{3}}{2} \theta + C$$
$$\theta = \tan^{-1}(u)$$
Thus, the final solution is:
$$\frac{4}{3} \frac{\sqrt{3}}{2} \tan^{-1} \left( \sqrt{\frac 4 3} \left(x+\frac 1 2 \right)\right) + C$$
However, according to online calculator integral-calculator, the answer is:
$$\frac 2 {\sqrt 3} \tan^{-1} \left( \frac{2x+1}{\sqrt 3} \right)+C$$
Any indication as to where my mistake falls would be very beneficial.

Your only mistake appears to be a failure to notice that $\displaystyle \frac{4}{3} \frac{\sqrt{3}}{2} \tan^{-1} \left( \sqrt{\frac 4 3} \left(x+\frac 1 2 \right)\right) $ is exactly the same thing as $ \displaystyle \frac 2 {\sqrt 3} \tan^{-1} \left( \frac{2x+1}{\sqrt 3} \right).$
First you have
$$
\frac 4 3 \cdot \frac{\sqrt 3} 2 = \frac{4\sqrt 3}{\sqrt 3\sqrt 3 \cdot 2} = \frac 2 {\sqrt 3}.
$$
And then
$$
\sqrt{\frac 4 3} \left( x + \frac 1 2 \right) = \frac 2 {\sqrt 3} \left( x + \frac 1 2 \right) = \frac 1 {\sqrt 3} (2x+1).
$$
