Calculate the integral: $ \int_{0}^{2} \! \sqrt{1+x^3}\,dx $ I need to calculate the integral:
$
\int_{0}^{2} \! \sqrt{1+x^{3}}\,dx 
$
I tried using variable substitution with $x^{3}+1=u$, however I could not solve it. Any idea how to solve this problem?
I have consulted with a software, and the answer is approximately 3,but I could not reach that answer.
 A: It is unlikely that the integral has a closed form in elementary functions.
By Chebyshev, the integral $x^p(a+bx^r)^q$ with $a,b\in\mathbb{R}^+$ has an elementary antiderivative if and only if at least one of $\displaystyle q,\frac{p+1}{r},\left(\frac{p+1}{r}+q\right)$ is an integer. (Ref 1, Ref 2 p. 10) In your case, $a,b=1,p=0,q=\frac12,r=3$ and none of $\displaystyle \frac12,\frac13,\frac56$ is an integer, so $\sqrt{1+x^3}$ has a nonelementary antiderivative. See also: Mathworld: Chebyshev integral. However, this doesn't directly imply that the definite integral doesn't have a closed form in elementary functions since nonelementary antiderivatives can be made into elementary expressions with appropriate bounds, e.g., $\displaystyle \int_0^1 \ln t\ln(1-t)\,\mathrm{d}t$.
The antiderivative does have an expression in terms of the elliptic integral of the first kind, $F(x|m)$ with parameter $m=k^2$ so with Mathematica and some algebraic simplification we can arrive at 
$$I=\frac{12}5+\frac{3^{3/4}(\sqrt{3}+i)}{10}\left(\sqrt{2}e^{-\pi i/3}(1+i)y^+-2e^{-\pi i/12}y^-\right)=3.2413092632$$
where
$$y^\pm=-F\left(\sin^{-1}\left(3^{\pm1/4}e^{11\pi i/12}\right)\left| \frac12+\frac{\sqrt{3}}2i\right.\right)\approx(1.78- 0.48i), (0.89- 0.24i)
$$
The values yield no results in the inverse symbolic calculator. 
A: As said in comments, this is a very difficult problem and I suppose a typo in the radical.
Anyway, you can approximate the value dividing the interval and use the fact that around $x=a$, by Taylor, we have
$$\sqrt{1+x^3}=\left(a^3+1\right)^{1/2}+\frac{3 a^2 }{2 \left(a^3+1\right)^{1/2}}(x-a)+\frac{3 a\left(a^3+4 \right)
   }{8 \left(a^3+1\right)^{3/2}}(x-a)^2+O\left((x-a)^3\right)$$ which does not make any problem.
Doing it for $4$ intervals gives a quite decent result.
A: By MVT
$$1(2-0) \le I \le 3 (2-0) \implies 2\le  I \le 6.$$
Next $I$ is ;ess than the area of trapezium whose corners are the points $(0,0), (0,1), (2,3), (0,2).$ So $I<4$.
By dividing  the domain in to $(0,1)$ and $(1,2)$
considering the areas of rectangles we get 
$$1+\sqrt{2} < I < 3+\sqrt{2} \implies  2.414 <I < 4.414$$ Numerically $I=3.2413..$
Also $I =2~ ~_2F_1[-1/2,1/3; 4/3,-8]$, where $~_2F_1[a,b;c,z]$
is Gauss hyper geometric function:
https://en.wikipedia.org/wiki/Hypergeometric_function
