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While simplifying a big integral I had to calculate the following two integrals

$$\int_{0}^{1}\frac{x\space \tan^{-1}(x)}{x^2+3}, \int_{0}^{1}\frac{x\space \tan^{-1}(x)}{3x^2+1}$$

$$I=\int_{0}^{1}\frac{x\space \tan^{-1}(x)}{x^2+3}$$

$$\implies I=\int_{0}^{1}\frac{x}{x^2+3}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1} x^{2n-1} dx$$

$$\implies I=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\int_{0}^{1}\frac{x^{2n}}{x^2+3} dx$$

Doing same thing with the second integral, we get

$$J=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\int_{0}^{1}\frac{x^{2n}}{3x^2+1} dx$$

But I am stuck at those 2 integrals

$$\int_{0}^{1}\frac{x^{2n}}{x^2+3} dx, \int_{0}^{1}\frac{x^{2n}}{3x^2+1} dx$$

Since they are almost the same so I think that doing one is enough to evaluate the other integral( Or Correct me If I'm wrong).

Thank you for your help!

K.defaoite
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User 1207
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    By long division, the integrand will be split in a polynomial of degree $n-2$ and a remainder term $\dfrac 1{x^3+3}$. –  Jun 29 '21 at 12:33

3 Answers3

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Let $$I_n = \int_{0}^{1}\frac{x^{2n}}{x^2+3}dx$$ Then we have the following recursion: $$I_{n+1}+3I_n = \int_0^1x^{2n}dx=\frac1{2n+1}$$ Use it to get a general form.

Martund
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Let $ n\in\mathbb{N}^{*} $. Using the geometric sum formula, we have for all $ x\in\mathbb{R} $ : $$ \sum_{k=0}^{n-1}{\frac{\left(-1\right)^{k}}{3^{k}}x^{2k}}=\frac{1-\frac{\left(-1\right)^{n}}{3^{n}}x^{2n}}{1+\frac{x^{2}}{3}}=\frac{3}{x^{2}+3}-\frac{\left(-1\right)^{n}}{3^{n-1}}\frac{x^{2n}}{x^{2}+3} $$

Thus, for all $ x\in\mathbb{R} $ : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x^{2n}}{x^{2}+3}=\frac{\left(-3\right)^{n}}{x^{2}+3}+\sum_{k=0}^{n-1}{\left(-3\right)^{n-1-k}x^{2k}}\end{array}$} $$

Hence : \begin{aligned} \int_{0}^{1}{\frac{x^{2n}}{x^{2}+3}\,\mathrm{d}x}&=\left(-3\right)^{n}\int_{0}^{1}{\frac{\mathrm{d}x}{x^{2}+3}}+\sum_{k=0}^{n-1}{\left(-3\right)^{n-1-k}\int_{0}^{1}{x^{2k}\,\mathrm{d}x}}\\ &=\left(-3\right)^{n}\frac{\pi}{6\sqrt{3}}+\sum_{k=0}^{n-1}{\frac{\left(-3\right)^{n-1-k}}{2k+1}} \end{aligned}

CHAMSI
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I think we can get a closed form. For that we simply solve

$$\Delta I_n +4 I_n = \frac{1}{2n+1},\ I_0 = \frac{\pi}{3\sqrt{3}}$$ (credit of @Martund) with which we obtain

$$I_n = (-3)^n\left(C-\frac{1}{3}\sum\left(-\frac{1}{3}\right)^{n}\frac{1}{2n+1}\right)=(-3)^n\left(C-\frac{1}{3}\int^{-\frac{1}{3}} \frac{t^{2n}}{t^2-1}dt\right)$$ and so

$$C=\frac{\pi}{3\sqrt{3}}+\frac{\ln\left(\frac{4}{3}\right)-\ln\left(\frac{2}{3}\right)}{6}=\frac{\pi}{3\sqrt{3}}+\frac{\ln(2)}{6}$$ thus

$$I_n =(-3)^n\left(\frac{\pi}{3\sqrt{3}}+\frac{\ln(2)}{6}+\left(-\frac{1}{3}\right)^{2n+2}\frac{{_2}F_1\left(1,n+\frac{1}{2},n+\frac{3}{2};\frac{1}{9}\right)}{2n+1} \right)$$

The hypergeometric series above should always have a closed form:

enter image description here

Alexander Conrad
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