I recently came across this integral $\int\sqrt{x^5+2}\; dx$. From Wolframalpha i can see that it has a closed form. how does one get to that closed form? what techniques should i approach?
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3$\begingroup$ Yeah but i dont have pro. And also how can i know what to do? To solve theese normaly $\endgroup$– AndersJensenApr 10, 2020 at 22:01
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4$\begingroup$ Think you should read the problem again :) $\endgroup$– AndersJensenApr 10, 2020 at 22:05
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4$\begingroup$ The solution is in terms of a hypergeometric function which should indicate that a series approach is being used. $\endgroup$– Cameron WilliamsApr 10, 2020 at 22:15
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$\begingroup$ The thing im woundering about is how do i find that series? what should i use to find it etc? and get to that final solution. $\endgroup$– AndersJensenApr 10, 2020 at 22:16
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$\begingroup$ You could find the general term in the Taylor Series expansion of $\sqrt{x^5+2}$ then integrate that. $\endgroup$– sammy gerbilApr 10, 2020 at 22:18
1 Answer
As said in comments, using Taylor or the binomial theorem, we have $$\sqrt{x^5+2}=-\frac{1}{\sqrt{2 \pi }}\sum_{n=0}^\infty(-1)^{n}\frac{ \left(n-\frac{3}{2}\right)!}{ 2^n\, n!} x^{5 n}$$ $$\int \sqrt{x^5+2}\,dx=-\frac{1}{\sqrt{2 \pi }}\sum_{n=0}^\infty(-1)^{n}\frac{ \left(n-\frac{3}{2}\right)!}{ 2^n\,(5n+1)\, n!} x^{5 n+1}$$ Computing the infinite summation $$S=x\sqrt{2} \,\, _2F_1\left(-\frac{1}{2},\frac{1}{5};\frac{6}{5};-\frac{x^5}{2}\right)$$ which looks simpler that the result from Wolfram Alpha but which numerically does not agree with it. However, the formula given here matches the results obtained by numerical integration.
What happens ? That is the question !
Edit
If I use the result given by Wolfram Alpha, $$\int \sqrt{x^5+2}\,dx=\frac{1}{7} x \left(5 \sqrt{2} \, _2F_1\left(\frac{1}{5},\frac{1}{2};\frac{6}{5};-\frac{x^5}{2}\right)+2 \sqrt{x^5+2}\right)$$ and differentiate is, hoping that I am not mistaken, the result is $$\frac{5 x^5}{2 \sqrt{x^5+2}}$$
On Wolfram Cloud, I obtained the result I gave.
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$\begingroup$ Thanks for the answer. What im really trying to understand by all this is how to go from $\int \sqrt{x^5+2}\,dx=-\frac{1}{\sqrt{2 \pi }}\sum_{n=0}^\infty(-1)^{n}\frac{ \left(n-\frac{3}{2}\right)!}{ 2^n\,(5n+1)\, n!} x^{5 n+1}$ and then to $S=x\sqrt{2} \,\, _2F_1\left(-\frac{1}{2},\frac{1}{5};\frac{6}{5};-\frac{x^5}{2}\right)$ $\endgroup$ Apr 11, 2020 at 9:44