# How can I solve this weird integral?

a friend of mine asked me this question, given by a book with many difficult math problems. And then, I saw that is quite interesting to start by trying to evaluate the problem by the substitution of $x^4+1$, so we can call it u. But, unfortunately, this couldn't help me, since the integral just got harder, as I found a new integral in terms of u, that has the annoying $4x$ messing up with this possible solution.

So, finally, after failing with Wolfram Alpha, Symbolab and others, I came here to ask you guys, how can I solve it? $$\int _1^3\:\sqrt{x^4+1} \ \ dx$$

Also, it even gets more difficult when he asks: What is the value of x, since $$\int _1^3\:\sqrt{x^4+1} \ \ dx\ge \frac{2x}{3}$$

Thank you all for spending some time analysing my question!

• Be careful with your second integral: the $x$ on the right-hand side is completely unrelated to the $x$s on the left-hand side! (I'm not sure your second expression is meaningful, actually.) – Patrick Stevens Mar 14 '17 at 22:17
• I sense a strog smell of elliptic integral. – Sangchul Lee Mar 14 '17 at 22:20
• Mathematica says the numerical answer is about 8.98005623469252857877121527304; it can't find a nice approximation to this number. – Patrick Stevens Mar 14 '17 at 22:24
• One usually tries $u$-substitution when the integrand contains a factor to correspond with $\mathrm{d}u$. It is possible a trignometric or other "special function" substitution will be required. – hardmath Mar 14 '17 at 22:25
• And the exact answer is, as simply as Mathematica can make it, $\frac{1}{3} \sqrt{2} \left(-1+3 \sqrt{41}+(1+i) F\left(\left.i \sinh ^{-1}\left(\sqrt[4]{-1}\right)\right|-1\right)-(1+i) F\left(\left.i \sinh ^{-1}\left(3 \sqrt[4]{-1}\right)\right|-1\right)\right)$, where $F$ is the elliptic integral of the first kind. – Patrick Stevens Mar 14 '17 at 22:26

$\int_1^3\sqrt{x^4+1}~dx$
$=\int_1^3x^2\sqrt{1+\dfrac{1}{x^4}}~dx$
$=\int_1^3x^2\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)x^{4n}}~dx$
$=\int_1^3\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2-4n}}{4^n(n!)^2(1-2n)}~dx$
$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{3-4n}}{4^n(n!)^2(1-2n)(3-4n)}\right]_1^3$
$=\sum\limits_{n=0}^\infty\dfrac{27(-1)^n(2n)!}{324^n(n!)^2(2n-1)(4n-3)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(2n-1)(4n-3)}$