<s>[INST] How Can I calculate $\int_{0}^{1}\frac{x^{2n}}{x^2+3}dx$ 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!
 [/INST] 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:

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