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Integral by parts: $$ I = x\sin^{-1}\left(x^{15}\right) - \int\frac{15x^{15}}{\sqrt{1-x^{30}}}dx $$ then what? The answer by wolfram gives an answer contains hypergeometric ${}_2F_1$ function,because it has no elementary answer. The question I want to know is, how can we find the integral of $$ \frac{15x^{15}}{\sqrt{1-x^{30}}} $$ in terms of hypergeometric function?

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  • $\begingroup$ yes,it is the function you write. $\endgroup$ – 390625 Oct 31 '14 at 11:31
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    $\begingroup$ Not that I have tried, but I would guess you just substitute $t=-x^{30}$ in the Maclaurin series for $(1+t)^{-1/2}$, and then integrate termwise. $\endgroup$ – Hans Lundmark Oct 31 '14 at 14:01
  • $\begingroup$ @HansLundmark: Please post that as an answer. $\endgroup$ – Lucian Oct 31 '14 at 16:28
  • $\begingroup$ @Lucian: I could, but a better answer was just posted anyway, so I'll just leave it as a comment. $\endgroup$ – Hans Lundmark Oct 31 '14 at 16:58
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$\int\dfrac{15x^{15}}{\sqrt{1-x^{30}}}dx$

$=\int_0^x15t^{15}(1-t^{30})^{-\frac{1}{2}}~dt+C$

$=\int_0^{x^{30}}15t^\frac{1}{2}(1-t)^{-\frac{1}{2}}~d(t^\frac{1}{30})+C$

$=\dfrac{1}{2}\int_0^{x^{30}}t^{-\frac{7}{15}}(1-t)^{-\frac{1}{2}}~dt+C$

$=\dfrac{1}{2}\int_0^1(x^{30}t)^{-\frac{7}{15}}(1-x^{30}t)^{-\frac{1}{2}}~d(x^{30}t)+C$

$=\dfrac{x^{16}}{2}\int_0^1t^{-\frac{7}{15}}(1-x^{30}t)^{-\frac{1}{2}}~dt+C$

$=\dfrac{15x^{16}}{16}~_2F_1\left(\dfrac{1}{2},\dfrac{8}{15};\dfrac{23}{15};x^{30}\right)+C$

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