Evaluate the integral: I = integral (sqrt
Evaluate the integral: $$ I = \int \frac{\sqrt[3]x}{\sqrt{\sqrt{x+1}}}\,dx = \int x^{1/3}(x+1)^{-1/4}\,dx. $$
Explanation of Method
This is a binomial differential $\int x^m(a+bx^n)^p\,dx$ with $m=\tfrac13,\ n=1,\ p=-\tfrac14$. Chebyshev's theorem says such an integral is elementary only if $p$, $\tfrac{m+1}n$, or $\tfrac{m+1}n+p$ is an integer. None of the three is an integer here, so this integral genuinely has no elementary closed form — unlike Problem 273, where the difficulty was practical rather than a proven impossibility, this one is provably non-elementary. It can still be written exactly using the hypergeometric function.
- Check Chebyshev's integrability conditions.
- Expand $(1+x)^{-1/4}$ as a binomial series.
- Integrate term by term to identify the hypergeometric series.
Step 1 · Chebyshev's theorem
For $\int x^m(1+x)^p\,dx$ with $m=\tfrac13,\ p=-\tfrac14$: $p=-\tfrac14\notin\mathbb Z$; $\ \tfrac{m+1}1=\tfrac43\notin\mathbb Z$; $\ \tfrac{m+1}1+p=\tfrac43-\tfrac14=\tfrac{13}{12}\notin\mathbb Z$. Since none of the three quantities is an integer, the integral cannot be reduced to elementary functions by any substitution.
Step 2 · Binomial series expansion
For $|x|<1$, $(1+x)^{-1/4}=\displaystyle\sum_{n=0}^\infty \binom{-1/4}n x^n$, so:
$$ x^{1/3}(1+x)^{-1/4} = \sum_{n=0}^\infty \binom{-1/4}n x^{n+1/3}. $$
Step 3 · Integrate term by term
$$ I = \sum_{n=0}^\infty \binom{-1/4}n \frac{x^{n+4/3}}{n+4/3} + C. $$
This series is precisely the definition of a hypergeometric function: using $\binom{-1/4}n=\dfrac{(-1)^n(1/4)_n}{n!}$ and the identity $\displaystyle\int_0^xt^{b-1}(1+t)^{-a}\,dt=\frac{x^b}b\,{}_2F_1(a,b;b+1;-x)$ with $a=\tfrac14,\ b=\tfrac43$:
$$ I = \frac{3x^{4/3}}4\,{}_2F_1\!\left(\frac14,\frac43;\frac73;-x\right) + C. $$
Summary Table
| Chebyshev quantity | Value | Integer? |
|---|---|---|
| $p$ | $-1/4$ | No |
| $(m+1)/n$ | $4/3$ | No |
| $(m+1)/n+p$ | $13/12$ | No |
Additional Notes
- Flag for the reader: this differs from Problems 253 and 275 (where "impossible" claims turned out to be fixable errors) and from Problem 273 (where the difficulty was practical, not proven). Here, Chebyshev's theorem proves no elementary antiderivative exists, regardless of how cleverly one searches for a substitution.
- The series in Step 3 converges for $|x|<1$; for $x\geq1$ the hypergeometric function is defined by analytic continuation, and the closed form in the result box remains valid.
- For numerical evaluation over a specific interval, direct numerical integration is the practical approach.
Result
The integral has no elementary closed form; its exact value is: