Two different answers of same indefinite integral 
Evaluate $$\int \frac {dx}{x^2-x+1}$$

Method 1
$$\int \frac {dx}{x^2-x+1}=\frac {4}{3}\int \frac {dx}{1+\frac {4(x-1/2)^2}{3}}$$
Put $u=x-1/2$
Hence $$\frac 43\int \frac {dx}{1+\frac {4(x-1/2)^2}{3}} =\frac 43\int \frac {du}{1+\frac {4u^2}{3}}$$
And then I could use $$\int \frac {dx}{1+x^2}=\arctan x$$
Method 2
Let $$I=\int \frac {dx}{x^2-x+1}$$
Put $x=\frac 1y$
Hence $dx=\frac {-dy}{y^2}$
Hence $$I=\int \frac {\frac {-dy}{y^2}}{\frac {1}{y^2}-\frac 1y+1}=-\int \frac {dy}{y^2-y+1}=-\int \frac {dx}{x^2-x+1}=-I$$
Hence $$I=-I\Rightarrow I=0$$

Why am I getting two different answers?  I guess i am missing some link in method 2.


The is not a definite integral and $y$ is not a dummy variable.
$$-\int \frac {dy}{y^2-y+1}=-\int \frac {dx}{x^2-x+1}$$
is incorrect.

Indeed, we have $\displaystyle I=\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$.
$\displaystyle I=-\int\frac{dy}{y^2-y+1}$ implies that $\displaystyle \int\frac{dx}{x^2-x+1}+\int\frac{dy}{y^2-y+1}$ is a constant (not necessarily zero).
As $\displaystyle y=\frac{1}{x}$, it means that $\displaystyle \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{\frac{2}{x}-1}{\sqrt{3}}\right)$ is a constant.
Note: This constant is different when $x>0$ and $x<0$.
