<s>[INST] How do I integrate $\frac1{x^2+x+1}$? 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?
 [/INST] 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?
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