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I have to integrate $\int (t^2+7)^{\frac{1}{3}}$dt

I tried some trigonometric substitution and then rationalisation but didn't get anything

Please provide a hint on how to proceed with this question

Please help!!!

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    $\begingroup$ I don't think you can expect an elementary anti-derivative. Are you sure you're supposed to find one (by hand)? $\endgroup$
    – StackTD
    Jan 24 '18 at 15:29
  • $\begingroup$ Are you sure about the power $\frac 13$ ? This makes the problem very difficult (no closed form solution). I bet for $\frac 12$. $\endgroup$ Jan 24 '18 at 15:30
  • $\begingroup$ No it is 1/3 for sure $\endgroup$ Jan 24 '18 at 15:30
  • $\begingroup$ Do you want to bet ? I am sure that, once more, there is a typo in a textbook. $\endgroup$ Jan 24 '18 at 15:31
  • $\begingroup$ Any answer is appreciated@StackTD $\endgroup$ Jan 24 '18 at 15:32
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Setting $$t=\sqrt{7}\cosh x\implies dt =\sqrt{7}\sinh xdx$$

$$\int (t^2+7)^{\frac{1}{3}}dt = 7^{2/3}\int (\cosh^2x+1)^{\frac{1}{3}} \sinh x\, dx = 7^{2/3}\int \sinh^{5/3} x\, dx$$

this integral cannot be written as elementary function. It belongs to the class of hypergeometric functions

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  • $\begingroup$ I tried that , but power is 1/3 , this is not helping for 1/3 power $\endgroup$ Jan 24 '18 at 15:35
  • $\begingroup$ I bet this is the hint for exponent $\frac12$. $\endgroup$ Jan 24 '18 at 15:35
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    $\begingroup$ This is very fine for $\frac 12$ but $\frac 13$ leads to hypergeometric function, I guess. $\endgroup$ Jan 24 '18 at 15:36
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    $\begingroup$ @ClaudeLeibovici ya I just check it does not an obvious integral $\endgroup$
    – Guy Fsone
    Jan 24 '18 at 15:37
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    $\begingroup$ @Isham it will make it worse $\endgroup$
    – Guy Fsone
    Jan 24 '18 at 15:42
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Positioning your integral:

\begin{equation} \int \left(t^2 + 7 \right)^{\frac{1}{3}}\:dt = \int_0^t \left(w^2 + 7 \right)^{\frac{1}{3}}\:dw = \int_0^t \frac{1}{ \left(w^2 + 7 \right)^{\frac{1}{3}}}\:dw \end{equation}

We can employ the solution that I've spoken to here: \begin{equation} \int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}, \frac{1}{1 + ax^n} \right)\right] \end{equation}

Here $a = 7$, $m = -\frac{1}{3}$, $k = 0$, and $n = 2$. Thus,

\begin{align} \int \left(t^2 + 7 \right)^{\frac{1}{3}}\:dt &= \frac{1}{2}\cdot 7^{\frac{0 + 1}{2} --\frac{1}{3}}\left[B\left(-\frac{1}{3} - \frac{0 + 1}{2}, \frac{0 + 1}{2}\right) - B\left(-\frac{1}{3}- \frac{0 + 1}{2}, \frac{0 + 1}{2}, \frac{1}{1 + 7t^2} \right)\right] \\ &= \frac{7^{\frac{5}{6}}}{2}\left[B\left(-\frac{5}{6}, \frac{1}{2}\right) - B\left(-\frac{5}{6}, \frac{1}{2}, \frac{1}{1 + 7t^2} \right)\right] \end{align}

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