Is the integral $ \int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1} $ converges or diverges. Determine if the following integral diverges/converges, if it converges, is it absolutely or conditionally converges. 
$$
\int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1}
$$
What i tried:
In that interval we know that: 
$$
0 < x
$$
Therefore we can write: 
$$
\frac{1}{\sqrt{x^2-x+1} - 1} < \frac{1}{\sqrt{x^2-2x+1} - 1} = \frac{1}{\sqrt{(x-1)^2} - 1}
$$
$$
=  \frac{1}{x - 2} 
$$
Now lets try to calculate the integral, using limits, we get: 
$$
\lim_{t \to 2^-}\int_{1}^{t}\frac{1}{x-2} = ln(x-2)|_{1}^{t} = ln(t-2) - ln(1-2)
$$
I cant calculate this integral, namely i dont get any limit that is a number. 
I thought that maybe i will get to a form like this: 
$$
\int_{a}^{b}\frac{1}{(b-x)^\alpha}, \int_{a}^{b}\frac{1}{(x-a)^\alpha}
$$
But also, if i will, and conclude that $ \int \frac{1}{x - 2}$ diverges in that interval, i couldnt use the comparison test to conclude about the original function. 
Can i have a hint? 
Thank you. 
 A: HINT:
Note that we have
$$\frac1{\sqrt{x^2-x+1}-1} = \frac{\sqrt{x^2-x+1}+1}{x(x-1)}\tag 1$$
How does the right-hand side of $(1)$ behave as $x\to 1^+$?  Does the improper integral $\int_1^2 \frac1{x-1}\,dx$ converge?
A: A quick way to decide about convergence/divergence is the limit comparison test:
$$\lim_{x\to 1^+}\frac{\sqrt{x^2-x+1}-1}{x-1}= \left. \left(\sqrt{x^2-x+1}\right)'\right|_{x=1}$$ $$ =\left. \frac{2x-1}{2\sqrt{x^2-x+1}}\right|_{x=1} = \frac 12$$
Since $\int_1^2\frac{dx}{x-1}$ is divergent, $\int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1}$ must be divergent, as well.
A: It diverges.
Multiplying by the conjugate:
$\frac{1}{\sqrt{x^2-x+1}-1} = \frac{\sqrt{x^2-x+1}+1}{x(x-1)}$.
A partial fraction decomposition yields:
$\frac{1}{x(x-1)}=\frac{1}{x}-\frac{1}{x-1}$ and therefore
$\frac{\sqrt{x^2-x+1}+1}{x(x-1)}=\frac{\sqrt{x^2-x+1}+1}{x}-\frac{\sqrt{x^2-x+1}+1}{x-1}$.
The right-hand side equals:
$\frac{\sqrt{x^2-x+1}}{x} - \frac{\sqrt{x^2-x+1}}{x-1} + \frac{1}{x} - \frac{1}{x-1}$
Also, remark that $\displaystyle \int_{1}^{2} \frac{\sqrt{x^2-x+1}}{x}dx < \infty$ and $\displaystyle \int_{1}^{2} \frac{dx}{x} < \infty$, hence it suffices to show that $\displaystyle \int_{1}^{2} \frac{\sqrt{x^2-x+1}}{x-1} dx$ diverges. Note that $\displaystyle \int_{1}^{2} \frac{x^2-x+1}{x-1} dx \geq \displaystyle \int_{1}^{2} \frac{dx}{x-1}$.
A: First, do a change of coordinates: $y=x-1$. The integral then becomes
$$\int_1^2\frac{1}{\sqrt{x^2-x+1}-1}dx=\int_0^1\frac{1}{\sqrt{y^2+y+1}-1}dy$$
Now, expand using the Taylor series around $0$ to get
$$\frac{1}{\sqrt{y^2+y+1}-1}=\frac{2}{y}-\frac{3}{2}+\frac{15 y}{8}-\frac{33 y^2}{16}+\cdots>\frac{2}{y}-\frac{3}{2}$$
This implies
$$\int_0^1\frac{1}{\sqrt{y^2+y+1}-1}dy>\int_0^1\left(\frac{2}{y}-\frac{3}{2}\right)dy=-\frac{3}{2}+2\int_0^1\frac{1}{y}dy$$
However, this final integral diverges, implying that the original integral also diverges.
