calc the length of $y = x^2-1$ I want to know the lenght of the curce $y = x^2-1$. To do this I use the integral method and get the integral $$\int \sqrt{(2x)^2+1)}dx$$. Substitute $2x = u$ and then $u+\sqrt{u^2+1} = s$ gives $$\frac{s^4-1}{8s^2}+\frac{1}{2}\ln(s)$$
The answer should be $\sqrt{5}+\ln(2+\sqrt{5})$. So the above primitive is correct if we integrate from 0 to 1.
However I've two problems with this:
Why is the integration from 0 to 1 and not from -1 to 1, or 0 to 1 and times 2?
I fail to simplify $\frac{s^4-1}{8s^2}$ to $\sqrt{5}$ and the only way I know that it is correct is by numerically solve it with a calculator. This is not okay...
=== Edit ===
I'm looking for the length of the curve below the x-axis
 A: I wrote about this topic together with two colleagues some years ago. We also derived almost the same formula as you gave which can be found as $L_{kurve}$ at the beginning of page 2 in the following (Danish) article:
LMFK-article on parabola length
As you can see, your answer is off by a factor $2$. This accounts for both problems! The limits $x=-1$ and $x=1$ works and $\sqrt 5$ 'pops out'. The primitive function should be
$$
\frac{s^4-1}{16s^2}+\frac{\ln(s)}{4}
$$
Plugging in $s=2+\sqrt 5$ the first fraction above becomes (noting that $s^2=9+4\sqrt 5$)
$$
\frac{s^4-1}{16s^2}=\frac{s^2}{16}-\frac{1}{16s^2}=\frac{9+4\sqrt 5}{16}-\frac{1}{16(9+4\sqrt 5)}
$$
then if we scale the last fraction by $(9-4\sqrt 5)$ noting that $(9+4\sqrt 5)(9-4\sqrt 5)=1$ we obtain
$$
\frac{s^4-1}{16s^2}=\frac{9+4\sqrt 5}{16}-\frac{9-4\sqrt 5}{16}=\frac{\sqrt 5}{2}
$$
Doing the same steps for $s=-2+\sqrt 5$ multiplying the last fraction by $(9+4\sqrt 5)$ this time will lead to a result of $-\frac{\sqrt 5}{2}$ and subtracting these two results we have the $\sqrt 5$ you asked for. There is one minor detail left. Subtracting the fractions with $\ln(s)$ one gets
$$
\frac{\ln(2+\sqrt 5)}{4}-\frac{\ln(-2+\sqrt 5)}{4}=\frac{\ln\left((2+\sqrt 5)/(-2+\sqrt 5)\right)}{4}
$$
where one can rewrite $(2+\sqrt 5)/(-2+\sqrt 5)=(2+\sqrt 5)^2$. Then dividing $\ln\left((2+\sqrt 5)^2\right)$ by $2$ to take the square root inside the logarithm we get the final result (that differs from yours!!!)
$$
\sqrt 5+\frac{\ln(2+\sqrt 5)}{2}
$$
This result can be verified by Wolfram Alpha Computation 1 and Wolfram Alpha Computation 2. Note that in computation 2 you should see the Alternative Form nr. 2 where the expression I gave is found.
A: That's because you haven't specified the limits of integration. The length of the curve from $x_1=?$ to $x_2=?$. I believe in your case, the limits were $x_1=0,x_2=1$.
Edit: The indefinite integral is $$\int\sqrt{4x^2+1}dx=\dfrac{x}{2}\sqrt{4x^2+1}+\dfrac{1}{4}\sinh^{-1}(2x)+C$$
Since the curve is under the $x$-axis and it is symmetric, $$\int^1_{-1}\sqrt{4x^2+1}dx=2\int^1_{0}\sqrt{4x^2+1}dx$$
Indeed, $\sqrt{5}+\ln(2+\sqrt{5})=2\int^1_{0}\sqrt{4x^2+1}dx=2\left(\dfrac{1}{4}\left(2\sqrt{5}+\sinh^{-1}(2)\right)\right)$
