How do I integrate $\frac1{x^2+x+1}$?

Question

I have tried this:

$$\frac1{x^2+x+1} = \frac1{\left( (x+\frac12)^2+\frac34\right)}$$

Now $u = x+\frac12$

$$\frac1{ u^2+\frac34 }$$ Now multiply by $ \frac34$

$$\frac1{ \frac43 u^2 + 1}$$

Now put the $\frac43$ outside the integral

$$\frac34 \int \frac1{u^2+1}\,du=\frac34\arctan(u)=\frac34\arctan(x+1/2)$$

But the result is not the same result calculated by computers.

What did I do wrong?

Could someone please help me with this?

I don't know where my wrong calculation is. The way should be correct to get to the result.

Edit:

So now

$\int \frac{1}{x^2+x+1}=\int \frac{1}{(x+\frac{1}{2})^2}= \int \frac{4}{3} \frac{1}{\frac{4}{3}(x+\frac{1}{2})^2+1}$ u=x^2+1/2 $\int \frac{4}{3} \frac{1}{\frac{4}{3}(u)^2+1}$

$\int \frac{4}{3} \frac{1}{\frac{4u^2}{3}+1}$

$\int \frac{4}{3} \frac{1}{\frac{2u^2}{\sqrt{3}}+1}$

Now it is:

$\frac{4}{3} \int \frac{1}{\frac{2u^2}{\sqrt{3}}+1}$

$=\frac{4}{3} * arctan(2*(x^2+1/2)/(\sqrt{3}))$

Why is this still not the same as the computer calculated solution?

Answer 1

Note that $$\frac{1}{u^2+3/4}=\frac{1}{\frac34((2u/\sqrt 3)^2+1)}=\frac43 \frac{1}{v^2+1}$$

where $v=2u/\sqrt 3$. Can you finish now?