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Any tips on how to prove this result?

$$ \int \frac{x}{x^K + c} dx = \frac{x^2 {}_2F_1 \left(1,\frac{2}{K};\frac{K+2}{K};-\frac{x^K}{c} \right)}{2c}, $$

where $${}_2F_1 (a, b;c;z)$$ is the hypergeometric function.

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Assuming $K,c>0$ , we can use the geometric series to find for $|x|^K < c$ $$ \int \limits_0^x \frac{t}{t^K + c} \, \mathrm{d} t = \frac{1}{c} \sum \limits_{n=0}^\infty \left(-\frac{1}{c}\right)^n \int \limits_0^x t^{K n +1} \, \mathrm{d}t = \frac{x^2}{c} \sum \limits_{n=0}^\infty \frac{1}{Kn+2} \left(-\frac{x^K}{c}\right)^n \, .$$

Since $$ \frac{1}{K n +2} = \frac{1}{2} \frac{\frac{2}{K}}{\frac{2}{K} +n} = \frac{1}{2} \frac{\Gamma\left(\frac{2}{K}+1\right)}{\Gamma\left(\frac{2}{K}\right)} \frac{\Gamma\left(\frac{2}{K}+n\right)}{\Gamma\left(\frac{2}{K}+1+n\right)} \, ,$$ we get $$ \int \limits_0^x \frac{t}{t^K + c} \, \mathrm{d} t = \frac{x^2}{2 c} \sum \limits_{n=0}^\infty \frac{\Gamma(1+n)}{\Gamma(1)} \frac{\Gamma\left(\frac{2}{K}+n\right)}{\Gamma\left(\frac{2}{K}\right)} \frac{\Gamma\left(\frac{2}{K}+1\right)}{\Gamma\left(\frac{2}{K}+1+n\right)} \frac{\left(-x^K/c\right)^n}{n!} \, .$$ Now note that the sum on the right-hand side is exactly the definition of the hypergeometric function.

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