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I'm trying to find closed form of this integral $$\mathcal{I} = \int_0^1 \frac{\arctan{x}}{x^2+x+1} \, \mathrm{d}x$$

I tried it by using differentiation integral sign, partial fraction but nothing seems to work. I was thinking to rewrite the denominator as $$\mathcal{I} = \int_0^1 \frac{(1-x)\arctan{x}}{1-x^3} \, \mathrm{d}x$$ And after using geometric series I got $$\mathcal{I} = \int_0^1 \left( \sum_{n=0}^\infty x^{3n} \right)(1-x)\arctan{x} \, \mathrm{d}x$$

We can interchange integral sign and summation because $x^{3n}$ is uniformly convergent in $0$ to $1$,

$$\mathcal{I} = \sum_{n=0}^\infty \int_0^1(x^{3n}-x^{3n+1})\arctan{x} \, \mathrm{d}x$$

Now after doing this I am thinking I have made the problem more harder...

I feel that this integral should have a closed form as this integral has $$\int_0^1 \frac{\arctan{x}}{x^2+1} \, \mathrm{d}x = \frac{\pi^2}{32}$$

That's how I thought... Thank you very much!!

Ted Shifrin
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Lucky Chouhan
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    Here is a similar integral, that might give you some ideas: https://math.stackexchange.com/q/3105528/42969 – Martin R Aug 16 '23 at 16:42
  • @MartinR yeah, I am going to find that's closed form first. Thank you :) – Lucky Chouhan Aug 16 '23 at 16:44
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    Your analogous integral is quite trivial, as it’s $\int u,du$. That doesn’t get you very far unless you have a formula for $\arctan(x+c)$. – Ted Shifrin Aug 16 '23 at 16:51
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    @TedShifrin exactly, I am also thinking about integrals of the form $$\int_0^1 \frac{\arctan(x+a)}{x^2+b^2} , \mathrm{d}x$$ $\mathcal{I}$ can be transformed into this form I guess.. – Lucky Chouhan Aug 16 '23 at 16:53
  • Mathematica produces an analytic expression with $\text{PolyLog}$s. – user64494 Aug 16 '23 at 16:56
  • integral-calculator.com actually finds an antiderivative with products and sums of polylogarithms and complex logarithms including a step-by-step derivation. – Kevin Dietrich Aug 16 '23 at 17:25
  • @KevinDietrich The issue is that using the option to integrate from 0 to 1 yields a negative number, even though the integrand is always positive – Andrei Aug 16 '23 at 17:27
  • The same integral is evaluated here, but over $[0,\infty)$ – user170231 Aug 16 '23 at 17:37
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    Maple does it using dilog. This comes from writing $$\arctan ! \left(x \right) = (\ln(1-ix) - \ln(1+ix)) i/2$$ and expanding $1/(x^2+x+1)$ in partial fractions. In general, $$\int \frac{\ln(x-a)}{x-b}; dx = \text{dilog}\left(\frac{x-b}{a-b}\right) + \ln(x-a) \ln\left(\frac{x-b}{a-b}\right) + C$$ – Robert Israel Aug 16 '23 at 17:44
  • @KevinDietrich Isn't it returning a horrifying answer... what else you think about it? – Lucky Chouhan Aug 16 '23 at 17:45
  • @RobertIsrael but $x^2+x+1$ have imaginary roots...... we can write it as $$x^2+x+1 = \left( x + \frac{1}{2} - \frac{\sqrt{3}i}{2} \right)\left( x + \frac{1}{2} + \frac{\sqrt{3}i}{2} \right)$$ – Lucky Chouhan Aug 16 '23 at 17:49
  • @LuckyChouhan Complex, not imaginary. Yes, $a$ and $b$ can be complex numbers. – Robert Israel Aug 17 '23 at 00:31

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