Calculate an integral using euler substituition.

[MathematicalPhysicist]

i was asked to calculate the integral:
$$\int\frac{dx}{x+\sqrt{x^2-x+1}}$$ by using euler substituition (i.e, finding a line which intersects sqrt(x^2-x+1) through one point and then the equation of the line will be y-y0=t(x-x0) where (x0,y0) is one point of intersection, and then substituing x for a rational function of t.
i did so in this particular example but i got stuck.
so i tried to recheck it through mathworld's integrator, but it states there isn't such formula for this integral, is this correct?
obviously if it's integrator, so it must be. (-:

 

[VietDao29]

Well, I'd do it this way, let t, such that:
$$\sqrt{x ^ 2 - x + 1} = t - x \quad (1)$$ or stated differently $$t = x + \sqrt{x ^ 2 - x + 1}$$

Now, we'll square both sides of (1):
$$\Rightarrow x ^ 2 - x + 1 = t ^ 2 - 2xt + x ^ 2$$
$$\Rightarrow - x + 1 = t ^ 2 - 2xt$$

Isolate x's to one side, and we'll try to express x in terms of t:
$$\Rightarrow - x + 2xt = t ^ 2 - 1$$

$$\Rightarrow x (2t - 1) = t ^ 2 - 1$$

$$\Rightarrow x = \frac{t ^ 2 - 1}{2t - 1} \quad (2)$$

Take the differential of both sides yields:

$$\Rightarrow dx = \frac{2 t ^ 2 -2t + 2}{(2t - 1) ^ 2} dt = 2 \frac{t ^ 2 - t + 1}{(2t - 1) ^ 2} dt \quad (3)$$

Now substitute (1), (2), and (3) to your original integral, we'll have:
$$\int \frac{dx}{x + \sqrt{x ^ 2 - x + 1}} = 2 \int \frac{\frac{t ^ 2 - t + 1}{(2t - 1) ^ 2} dt}{t} = 2 \int \frac{t ^ 2 - t + 1}{(2t - 1) ^ 2 t} dt$$
It looks much better than its original form, right?
Can you go from here? :)