Help to calculate $\int \frac{dx}{x-\sqrt{x^2-x+1}}$

Question

I have a problem with calculating this integer:

$\int \frac{dx}{x-\sqrt{x^2-x+1}}$

I found this after some calculations:

$\int \frac{dx}{x-\sqrt{x^2-x+1}} = \int 1dx + \int \frac{dx}{x-1} + \int \frac{\sqrt{x^2-x+1}}{x-1} = x + ln|x-1| + \int \frac{\sqrt{x^2-x+1}}{x-1}$

I have no ideas what to do with this part: $ \int \frac{\sqrt{x^2-x+1}}{x-1} $, but maybey my approach is wrong

Could someone give me a helping hand?

Answer 1

For the inetgral $I= \int \frac{dx}{x-\sqrt{x^2-x+1}} $, you can use Euler's substitution, let $\sqrt{x^2-x+1}=x+t$

$$\Rightarrow x=\frac{1-t^2}{1+2t},~~ dx=-\frac{2 (1 + t + t^2)}{(1 + 2 t)^2}dt$$

$$\Rightarrow I= \int \frac{2 (1 + t + t^2)}{t(1 + 2 t)^2}dt$$

Next, you can use partial fraction to solve it.

Answer 2

Some hints:

Complete the square of the expression under the square root.

Then perform the $u$ substitution and then trigonometric substitution.

Then perform tangent half angle substitution then $w=\tan\frac{\theta}{2}$ substitution.

Then apply partial fraction decomposition.

Side note: The answer is very big if you want to compute it in elementary functions.

The answer to the expression which you were unable to solve is $$\dfrac{\ln\left(\frac{\left|2\left(\sqrt{x^2-x+1}+x\right)-\sqrt{3}-1\right|}{\left|2x-1\right|}\right)-\ln\left(\frac{\left|2\sqrt{x^2-x+1}-2x-\sqrt{3}+1\right|}{\left|2x-1\right|}\right)}{2}-\ln\left(\dfrac{\left|\sqrt{x^2-x+1}+\left(\sqrt{3}+2\right)x-\sqrt{3}-1\right|}{\left|2x-1\right|}\right)+\ln\left(\dfrac{\left|\sqrt{x^2-x+1}+\left(\sqrt{3}-2\right)x-\sqrt{3}+1\right|}{\left|2x-1\right|}\right)+\sqrt{x^2-x+1}+C$$