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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?

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    $\begingroup$ Please use MathJax to render your math equations $\endgroup$ – Shubhrajit Bhattacharya Jul 30 '20 at 23:06
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    $\begingroup$ $\dfrac1{u^2+\frac34}$ is not the same as $\dfrac1{\frac43u^2+1}$, which is not the same as $\dfrac34\dfrac1{u^2+1}$ (try $u=0$, for example) $\endgroup$ – J. W. Tanner Jul 30 '20 at 23:27
  • $\begingroup$ @ShubhrajitBhattacharya how do I use mathjax here? I just know LaTex $\endgroup$ – Rapiz Jul 30 '20 at 23:58
  • $\begingroup$ @Rapiz go here math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Shubhrajit Bhattacharya Jul 31 '20 at 0:08
  • $\begingroup$ Also MathJax is almost same as LaTeX except some differences that you will come across if you continue to use MathJax $\endgroup$ – Shubhrajit Bhattacharya Jul 31 '20 at 0:10
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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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    $\begingroup$ Again we have $$\frac1{u^2+3/4}=\frac{4/3}{4/3}\frac1{u^2+3/4}=\frac{4/3}{4u^2/3+1}=\frac43 \frac1{(2u/\sqrt 3)^2+1}$$ $\endgroup$ – Mark Viola Jul 30 '20 at 23:17
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    $\begingroup$ @Rapiz Multiplying top and bottom by $4/3$ gives $\frac{4}{3} \frac{1}{(4/3) u^2+1}$. You can't just multiply the bottom alone by $4/3$, that would change the result. $\endgroup$ – Ian Jul 30 '20 at 23:18
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    $\begingroup$ @Rapiz No, because now you've changed the integrand, that's what I was trying to say. If you want to multiply the bottom by $4/3$ (so as to make the $3/4$ into a $1$ without factoring the way that Mark Viola did) then you have to do the same to the top. $\endgroup$ – Ian Jul 30 '20 at 23:22
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    $\begingroup$ Is the ratio $\frac{a}{b}=\frac{a}{(4/3)b}$? It isn't. But we can write $$\frac ab=\frac{(4/3)a}{(4/3)b}$$ $\endgroup$ – Mark Viola Jul 30 '20 at 23:23
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    $\begingroup$ You cannot simply multiply the denominator by $4/3$ and not do the same to the numerator. The integrand IS a ratio. It is $\frac1{u^2+3/4}$. So, here $a=1$ and $b=u^2+3/4$. Now how can we preserve equality? We can multiply by $1=\frac{4/3}{4/3}$. $\endgroup$ – Mark Viola Jul 30 '20 at 23:27

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