General Integral Question (not a specific problem)

[Asphyxiated]

Homework Statement

My book states that "The only way that we can show that the function $$f(x) = \sqrt {1 + x^{4}}$$ has an anti-derivative is to take a definite integral,

$$\int^{x}_{0} \sqrt {1 + t^{4}} dt$$

This is a new function that cannot be expressed in terms of algebraic, trigonometric or exponential functions without calculus."

What are they really saying here? I don't understand the point. Why is the anti-derivative of
$$f(x) = \sqrt {1 + x^{4}}$$ an integral? And why do we switch the variables from x to t with respect to t?

Second Part Of My Question

From my book:

"The Fundamental Theorem can also be used to find the derivative of a function which is defined as a definite integral with a variable limit of integration. This can be done without actually evaluating the integral."

Examples:

Let

$$y = \int^{2}_{x} \sqrt{1+t^{2}} dt$$

then

$$y = - \int^{x}_{2} \sqrt{1+t^{2}} dt$$

and

$$dy = -d( \int^{x}_{2} \sqrt{1+t^{2}}dt) = -\sqrt{1+x^{2}} dx$$

My question here arises from the same thing, why do they take a negative integral before taking a derivative and why does the derivative switch from t dt to x dx? The only reason I can see from the example is that x is the upper limit of the integral.. is that why?

A second example given:

Let

$$y = \int^{x^{2}+x}_{3} \frac {1} {t^{3}+1} dt$$

Let

$$u = x^{2}+x$$

Then

$$\frac {du}{dx} = (2x+1)$$ ... $$y = \int^{u}_{3} \frac {1}{t^{3}+1} dt$$

$$\frac {dy}{du} = \frac {1}{u^{3}+1}$$

then by the chain rule

$$\frac {dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \frac{1}{u^{3}+1} (2x+1) = \frac {2x +1} {(x^{2}+x)^{3} +1}$$

I have basically the same questions as before here, so any advice would be great.

The only thing that is after this in my book is questions for the chapter so I am kind of at a loss right now.

Thanks!

 

[Asphyxiated]

Actually never mind, I got it, thanks anyway!