Integral with trig substitution

[mbrmbrg]

The bit of the problem that I'm working on:
$$6\int\frac{dx}{x^2-x+1}$$

My work:
$$=6\int\frac{dx}{(x^2-x+\frac{1}{4})+1-\frac{3}{4}}$$
$$=6\int\frac{dx}{(x-\frac{1}{2})^2+\sqrt{\frac{3}{4}}^2}$$

let $$x-\frac{1}{2}=\sqrt{\frac{3}{4}}\tan\theta$$
so $$dx=\sqrt{\frac{3}{4}}\sec^2\theta d\theta$$

$$=6\int\frac{\sqrt{\frac{3}{4}}sec^2\theta d\theta}{\frac{3}{4}+\frac{3}{4}tan^2\theta}$$
$$=(6)(\frac{\sqrt{3}}{2})(\frac{4}{3})\int\frac{sec^2\theta d\theta}{1+tan^2\theta}$$
$$=4\sqrt{3}\int d\theta$$
$$=4\sqrt{3}\theta$$

My answer:
$$=4\sqrt{3}\arctan{\frac{2x-1}{\sqrt{3}}}$$

An integrator's answer:
$$\frac{2}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}$$

I don't think the two answers differ by a constant, but I can't find my error.

 

[AKG]

Your answer is 6 times greater than the integrator's. I assume you just didn't include the 6 when you put your integrand into the integrator since you can't write stuff before the integral when using that program. Try using the integrator again, bringing the 6 into the numerator of the integrand.

 

[mbrmbrg]

Thank you.