# Math Help - Fundamental Theorm

1. ## Fundamental Theorm

Find the derivative of the function.

I was watching a video on how to seperate them and they used 0. Can I do that with these two functions?

What I am talking about is breaking the original equation down to...

$\int_5x^0 frac_((u^2)-1)/((u^2)+1)$ $+\int_0^6x frac_((u^2)-1)/((u^2)+1)$

Or is that completely wrong?

Wow so I suck at making integrals...but the first one is 5x on the bottom and 0 on top. the second is 0 on bottom and 6x on top....sorry

2. ryan18:

Yes that's fine; that's called the Additive Property of Integrals. Are you deriving or integrating? To integrate that function use partial fractions.

3. Originally Posted by ryan18
Find the derivative of the function.

I was watching a video on how to seperate them and they used 0. Can I do that with these two functions?

What I am talking about is breaking the original equation down to...

$\int_5x^0 frac_((u^2)-1)/((u^2)+1)$ $+\int_0^6x frac_((u^2)-1)/((u^2)+1)$

Or is that completely wrong?

Wow so I suck at making integrals...but the first one is 5x on the bottom and 0 on top. the second is 0 on bottom and 6x on top....sorry

$\frac{u^2-1}{u^2+1}=1-\frac{2}{u^2+1}\Longrightarrow \int\limits^{6x}_{5x}\frac{u^2-1}{u^2+1}\,dx=\left[u-2\arctan u\right]^{6x}_{5x}$ $=6x-2\arctan 6x-5x+2\arctan 5x$ , and now make some algebraic order here and differentiate.

Tonio

4. Originally Posted by ryan18
Find the derivative of the function.

I was watching a video on how to seperate them and they used 0. Can I do that with these two functions?

What I am talking about is breaking the original equation down to...

$\int_5x^0 frac_((u^2)-1)/((u^2)+1)$ $+\int_0^6x frac_((u^2)-1)/((u^2)+1)$

Or is that completely wrong?

Wow so I suck at making integrals...but the first one is 5x on the bottom and 0 on top. the second is 0 on bottom and 6x on top....sorry

Yes you can use any constant at all; won't make any difference. But I will just use a constant C.

$g(x) = \int_{5x}^{c}\frac{u^2 - 1}{u^2 + 1}du + \int_{c}^{6x}\frac{u^2 - 1}{u^2 + 1}du$

but since the variable has to be as the upper limit, we'll put a minus sign in front of the first integral and switch the limits

$g(x) = -\int_{c}^{5x}\frac{u^2 - 1}{u^2 + 1}du + \int_{c}^{6x}\frac{u^2 - 1}{u^2 + 1}du$

Then the derivative will be

$\frac{dg(x)}{dx} = -5\frac{25x^2 - 1}{25x^2 + 1} + 6\frac{36x^2 - 1}{36x^2 + 1}$

Also, practice your LaTeX typing skills.. it'll come in handy =P