Can't find error in completing the square

Question

Now matter what I do I can't seem to find the error in my completion of the square... It's probably something obvious but I'm running out of ideas at this point.

I'm trying to complete : $ 3 x^{2}-2 x+4 $ in order to solve this integral : $ \int \frac{d x}{3 x^{2}-2 x+4} $

I get : $$ \left(x-\frac{1}{3}\right)^{2}+\frac{11}{9} $$

But by going through with the integral (and I am 100% sure of the integral calculation) I do not arrive at the correct answer: hence my 'basic' square completion is has to be wrong, but I really can't see how else I would do it. Here is how I went about doing it: $$ \left(x-\frac{2 x}{3}-\frac{1}{2}\right)^{2}+\frac{4}{3}-\frac{1}{9} $$

I am also open to any suggestions for better methods of course !

Answer 1

You have missed a factor $3$:$$3x^2-2x+3=3\left(\left(x-\frac13\right)^2+\frac{11}9\right).$$In fact\begin{align}3x^2-2x+4&=3\left(x^2-\frac23x+\frac43\right)\\&=3\left(\left(x-\frac13\right)^2-\frac1{3^2}+\frac43\right)\\&=3\left(\left(x-\frac13\right)^2+\frac{11}9\right).\end{align}

Answer 2

We have $$3x^2-2x+4=3\left(x^2-\frac{2}{3}x\right)+4$$

Note $$\left(\frac{b}{2}\right)^2=\left(\frac{-\frac{2}{3}}{2}\right)^2=\frac{4}{36}=\frac{1}{9}$$ Thus $$3\left(x^2-\frac{2}{3}x\right)+4=3\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+4-\frac{3}{9}$$ $$=3\left(x-\frac{1}{3}\right)^2+4-\frac{1}{3}=3\left(x-\frac{1}{3}\right)^2+\frac{12}{3}-\frac{1}{3}$$ $$=3\left(x-\frac{1}{3}\right)^2+\frac{11}{3}$$

Therefore $$I=\int\frac{dx}{3x^2-2x+4}=\int\frac{dx}{3\left(x-\frac{1}{3}\right)^2+\frac{11}{3}}=\frac{1}{3}\int\frac{dx}{\left(x-\frac{1}{3}\right)^2+\frac{11}{9}}$$ $$=\frac{1}{3}\int\frac{dx}{\frac{11}{9}\left(\frac{\left(x-\frac{1}{3}\right)^2}{\frac{11}{9}}+1\right)}=\frac{3}{11}\int\frac{dx}{\frac{\left(x-\frac{1}{3}\right)^2}{\frac{11}{9}}+1}=\frac{3}{11}\int\frac{dx}{\left(\frac{3\left(x-\frac{1}{3}\right)}{\sqrt{11}}\right)^2+1}$$

Now let $u=\frac{3\left(x-\frac{1}{3}\right)}{\sqrt{11}}\implies du=\frac{3}{\sqrt{11}}\space dx\implies dx=\frac{\sqrt{11}}{3}du$. Thus $$I=\frac{3}{11}\int\frac{\frac{\sqrt{11}}{3}}{u^2+1}du=\frac{\sqrt{11}}{11}\int\frac{du}{u^2+1}=\frac{1}{\sqrt{11}}\arctan(u)+C$$ $$I=\frac{1}{\sqrt{11}}\arctan\left(\frac{3\left(x-\frac{1}{3}\right)}{\sqrt{11}}\right)+C$$