Evaluate :$$\int_0^1 \frac{1}{\left(1+\sqrt{\frac{1}{x}-1}\right)(x^2-x-1) }\mathrm dx$$ How to evaluating this integral, I don't know how to do it, and any help is welcome .
I tried using wolframalpha ,but the result is too complicated.
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2 Answers
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Hint: $y=1-x$ in the integral $I$, then show that $$2I=\int_0^1 \frac{dx}{x^2-x-1}$$
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$\begingroup$ +1 Can you please add some detail? $\endgroup$– RaffaeleCommented Sep 6, 2017 at 8:41
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$\begingroup$ By summing both expressions of the same integral: $I=\int_0^1 \frac{\sqrt{1-y}}{(\sqrt{y}+\sqrt{1-y})(y^2-y-1)}\,dy$ and $I=\int_0^1 \frac{\sqrt{x}}{(\sqrt{x}+\sqrt{1-x})(x^2-x-1)}\,dx$ one gets a simpler form: $I=\frac{1}{2}\int_0^\infty \frac{dx}{x^2-x-1}$. Then, by changing $z=(2x-1)/\sqrt{5}$ reduces to $I=-\frac{1}{\sqrt{5}}\int_{-1/\sqrt{5}}^{1/\sqrt{5}}\frac{dz}{1-z^2}=-\frac{2}{\sqrt{5}}\tanh^{-1}\frac{1}{\sqrt{5}}$, as given in Connor Evans' answer. $\endgroup$ Commented Sep 6, 2017 at 9:17
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$\begingroup$ @Paul Enta Thank you very much for your answer. $\endgroup$– JamesJCommented Sep 6, 2017 at 10:39
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If you are just looking for the answer its
$$ -\frac{2}{5}\sqrt{5}\tanh^{-1}(\frac{1}{5}\sqrt{5}) $$
(by Maple)
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2$\begingroup$ it kinda reduces to $-2\tanh ^{-1}(\frac15)$ $\endgroup$ Commented Sep 5, 2017 at 20:52
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$\begingroup$ Not according to Maple buddy $\endgroup$– user382669Commented Sep 5, 2017 at 20:58
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1$\begingroup$ Well, you did edit the post, buddy $\endgroup$ Commented Sep 5, 2017 at 22:11