Integral with trig substitution

[mbrmbrg]

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

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

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

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

My answer:
[tex]=4\sqrt{3}\arctan{\frac{2x-1}{\sqrt{3}}}[/tex]

An integrator's answer:
[tex]\frac{2}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}[/tex]

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.