TITLE: Prove the integral $\int_{0}^{1}\frac{x^{2n}}{x^{2}+a^{2}}dx$ has the recurrence relation $K_{n}+a^{2}K_{n-1}=\frac{1}{2n-1}$
QUESTION [1 upvotes]: Prove the integral $$\int_{0}^{1}\frac{x^{2n}}{x^{2}+a^{2}}dx$$
Has the following recurrence relation $$K_{n}+a^{2}K_{n-1}=\frac{1}{2n-1}\tag{$n \ge 1$}$$

I used integration by parts, let $x^{2n}=f'(x)$ and $\frac{1}{x^{2}+a^{2}}=g(x)$, then from $$\int_{ }^{ }f'\left(x\right)g\left(x\right)dx=f\left(x\right)g\left(x\right)dx-\int_{ }^{ }f\left(x\right)g'\left(x\right)dx$$
We have $$K_n=\frac{1}{2n+1}\cdot\frac{1}{1+a^{2}}+\frac{2}{2n+1}\int_{0}^{1}\frac{x^{\left(2n+2\right)}}{\left(x^{2}+a^{2}\right)^{2}}dx=$$
But I don't see how this can help to get the desired relation.

REPLY [1 votes]: You have\begin{align}K_n+a^2K_{n-1}&=\int_0^1\frac{x^{2n}}{x^2+a^2}+\frac{a^2x^{2n-2}}{x^2+a^2}\,\mathrm dx\\&=\int_0^1x^{2n-2}\,\mathrm dx\\&=\frac1{2n-1}.\end{align}
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