Why is $f(x) = \frac{x^{n-1}}{(1+x^{2})^{k}}$ Lebesgue-integrable over $(0,\infty)$ for all $k > \frac{n}{2}$?

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I want to show that, for all $n \in \mathbb{N}$ and for all $k > \frac{n}{2}$ $$f(x) = \frac{x^{n-1}}{(1+x^{2})^{k}}$$

is Lebesgue-integrable over $X = (0,\infty)$. I was able to show that $f(x)$ is Lebesgue-integrable over all $(0,s)$ where $s>0$. However, I couldn't find a proof for the desired set. I was trying to find a Lebesgue-integrable function $g$ such that $0\le f \le g$ so $$\int_{X} f dm_1 \le \int_{X} g dm_1 < +\infty$$

Therefore, f is Lebesgue-integrable. But all of my attempts were unsuccessful :'c.

Can you give me a hint o tell me if I am wrong?

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You really want to split the interval into $ [0, 1] $ and $ [1, \infty] $. It's easy to see that $ f $ is integrable on $ [0,1] $ because it's bounded. For $ [1, \infty] $, find a constant $ C $ such that you can compare $ f(x) $ to $ C x^{n-1-2k} $. If this seems non-intuitive to you consider my intuition: "for large values of $ x $, the $ + 1 $ becomes insignificant."