How to calculate $\int\frac{x}{x^2-x+1}\, dx$? $$\int \frac{x}{x^2-x+1}\,  dx = \int \frac{x}{(x-\frac 1 2)^{2} + \frac 3 4}\,  dx = \int \frac{x}{(x-\frac 1 2)^2 + (\frac {\sqrt{3}} {2})^2}$$
Substitute $u= \frac{2x-1}{\sqrt{3}}, du=\frac{2}{\sqrt{3}}dx$:
$$\frac {\sqrt{3}} 2 \int \frac{\frac{\sqrt{3}} {2}u + \frac 1 2}{(\frac{\sqrt{3}}{2}u)^2+(\frac {\sqrt{3}}{2})^2} = \int \frac{u}{u^2+1}du  + \frac{1}{\sqrt{3}}\int\frac 1 {u^2+1}du.$$
This gives $$\frac 1 2\log({u^2+1})+\frac{1}{\sqrt{3}}\arctan{u}.$$ 
Substituting back in x yields $$\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3)+\frac 1 {\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$
However, according to Wolfram Alpha, the integral should evaluate to $$\frac 1 2 \log(x^2-x+1)+\frac{1}{\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$After working some time on the integral, I know how to reach this solution, but I don't understand why my first attempt didn't arrive at the correct answer. Do you see where I went wrong? 
Thank you for any help!

$$\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3)+C$$
Can be written as 
$$\frac 1 2 \log (\frac 4 3) +\frac 1 2\log( x^2-x+ 1)+C $$
Where first term can be added to the constant
$$\frac 1 2 \log (\frac 4 3) +C     $$ write it as new constant $C_1$
