Does it?

1. Nov 12, 2007

nicodoggie

Can the integral be found?

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

Last edited: Nov 12, 2007
2. Nov 12, 2007

HallsofIvy

Staff Emeritus
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.

3. Nov 12, 2007

nicodoggie

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

4. Nov 12, 2007

Kummer

Elliptic Functions.

5. Nov 13, 2007

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

6. Nov 13, 2007

PowerIso

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

7. Nov 13, 2007

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!