Exploring the Solvability of Integrals: \int \sqrt{u^{4}+1} dx

[nicodoggie]

Can the integral be found?

Does the integral $$\int \sqrt{u^{4}+1} dx$$ have a solution?

 

[HallsofIvy]

What exactly are you asking? In the integral you give, you have a function of u integrated with respect to x! Is u some function of x or is that simply a typo?
If you meant $\int\sqrt{x^4+1}dx$, then since $\sqrt{x^4+1}$ is continuous, yes, there certainly exist a function having that as its derivative- it has an anti-derivative.

If, however, you are asking whether that anti-derivative can be written in terms of "elementary functions", no it cannot.

 

[nicodoggie]

I'm sorry, yes that is what I meant. I thought not. Thanks a lot!

 

[Kummer]

Elliptic Functions.

 

[nicodoggie]

Actually, I encountered the integral in a contest. The question was:

$$\frac{d^{2}}{(dx)^{2}}\int^{x}_{0}\left(\int^{sin t}_{1}\sqrt{u^{4}+1} du\right) dt}$$

I wasn't so sure how to solve it, but what I did was:
(I'm not even sure this is proper use of the fundamental theorem of the Calculus...)

$$=\frac{d^{2}}{(dx)^{2}} \int \left(\int^{sin x}_{1}\sqrt{u^{4}+1}du+\int^{sin 0}_{1}\sqrt{u^{4}+1}du\right)dx$$

$$=\frac{d}{dx} \left(\int^{sin x}_{1}\sqrt{u^{4}+1}du+\int^{sin 0}_{1}\sqrt{u^{4}+1}du\right)dx$$

$$=\frac{d}{dx}\left(\int \sqrt{sin^{4}x+1}dx-\int \sqrt{2}dx - \left(\int dx-\int \sqrt{2}dx\right) \right)$$

$$=\frac{d}{dx}\left(\int \sqrt{sin^{4}x+1}dx-\int dx\right)$$
$$=\underline{\sqrt{sin^{4}x+1}-1}$$

But they said the answer was $$cos x \sqrt{sin^{4}x+1}$$

Could anyone please tell me what I did wrong? Thanks

 

[PowerIso]

I think when you replaced the u with sinx you forgot to also do the chain rule. That would get you the cosx.

 

[nicodoggie]

oh yeah... makes sense. We were never really taught this stuff in class, (I don't think they deemed it necessary for IT majors to learn theory.) Thanks a lot!