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How to prove $\int_0^1 \frac{1+x^{30}}{1+x^{60}} dx = 1 + \frac{c}{31}$, where $0 \lt c \lt 1$

How can I prove the estimate $$ 1 \lt \int_0^1 \frac{1+x^{30}}{1+x^{60}} \ \mathrm{d}x \lt 1 + \frac{1}{30}?$$ Of course, the lower bound is pretty obvious. I realize this looks like a homework problem, but it's not (it's actually an old qualifying exam question). I think it is possible to use contour integration in the complex plane to get an exact expression for the integral, but this is pretty complicated and hopefully there is a way to estimate the integral without evaluating it.

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marked as duplicate by David Mitra, Argon, Marvis, Justin Campbell, Did Aug 4 '12 at 21:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See here. – David Mitra Aug 4 '12 at 19:58
Oops, I didn't see the duplicate. I'm voting to close as well. – Justin Campbell Aug 4 '12 at 20:35

$$1 < \dfrac{1+x^{30}}{1+x^{60}} < 1+x^{30} \,\,\,\,\, \forall x \in (0,1)$$

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I think this is the best way one may think of. (+1) – user 1618033 Aug 9 '12 at 13:23

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