Arc length from $\exp(x)$ , $x$ from $0$ to $t$ How do I find $t$ such that arc length of $\exp(x)$ from $x=0$ to $t$ is $2\pi$?
https://en.wikipedia.org/wiki/Arc_length
I know that it would be equal to $$\int_0^{t} \sqrt{\exp(2x)+1}\ dx$$
 A: Well, first of all, let's look at the integral:
$$\mathcal{I}_{\space\text{n}}\left(t\right):=\int_0^t\sqrt{1+\exp\left(\text{n}\cdot x\right)}\space\text{d}x\tag1$$
Now, let's do a couple substitutions:


*

*Substitute $\text{u}:=\text{n}\cdot x$

*Substitute $\text{s}:=\exp\left(\text{u}\right)$

*Substitute $\text{p}:=\sqrt{1+\text{s}}$


So, then we get:
$$\mathcal{I}_{\space\text{n}}\left(t\right)=\frac{2}{\text{n}}\cdot\int_\sqrt{2}^{\sqrt{1+\exp\left(\text{n}\cdot t\right)}}\frac{\text{p}^2}{\text{p}^2-1}\space\text{d}\text{p}\tag2$$
Using long divison, we can write:
$$\frac{\text{p}^2}{\text{p}^2-1}=1+\frac{1}{2}\cdot\frac{1}{\text{p}-1}-\frac{1}{2}\cdot\frac{1}{\text{p}+1}\tag3$$
Now, let's do a couple substitutions again:


*

*Substitute $\text{w}:=\text{p}+1$

*Substitute $\text{v}:=\text{p}-1$


So, we get:
$$\mathcal{I}_{\space\text{n}}\left(t\right)=\frac{2}{\text{n}}\cdot\int_\sqrt{2}^{\sqrt{1+\exp\left(\text{n}\cdot t\right)}}1\space\text{d}\text{p}+\frac{1}{\text{n}}\cdot\int_{\sqrt{2}-1}^{\sqrt{1+\exp\left(\text{n}\cdot t\right)}-1}\frac{1}{\text{v}}\space\text{d}\text{v}-\frac{1}{\text{n}}\cdot\int_{1+\sqrt{2}}^{1+\sqrt{1+\exp\left(\text{n}\cdot t\right)}}\frac{1}{\text{w}}\space\text{d}\text{w}=$$
$$\frac{2}{\text{n}}\cdot\left(\sqrt{1+\exp\left(\text{n}\cdot t\right)}-\sqrt{2}\right)+\frac{1}{\text{n}}\cdot\ln\left|\frac{\sqrt{1+\exp\left(\text{n}\cdot t\right)}-1}{\sqrt{2}-1}\right|-\frac{1}{\text{n}}\cdot\ln\left|\frac{1+\sqrt{1+\exp\left(\text{n}\cdot t\right)}}{1+\sqrt{2}}\right|=$$
$$\frac{2}{\text{n}}\cdot\left(\sqrt{1+\exp\left(\text{n}\cdot t\right)}-\sqrt{2}\right)+\frac{1}{\text{n}}\cdot\left(\ln\left(3+2\sqrt{2}\right)+\ln\left|\frac{\sqrt{1+\exp\left(\text{n}\cdot t\right)}-1}{\sqrt{1+\exp\left(\text{n}\cdot t\right)}+1}\right|\right)\tag4$$

For your case, Let $\text{n}=2$, so we want to solve:
$$\frac{2}{2}\cdot\left(\sqrt{1+\exp\left(2\cdot t\right)}-\sqrt{2}\right)+\frac{1}{2}\cdot\left(\ln\left(3+2\sqrt{2}\right)+\ln\left|\frac{\sqrt{1+\exp\left(2\cdot t\right)}-1}{\sqrt{1+\exp\left(2\cdot t\right)}+1}\right|\right)=2\pi\tag5$$
Using Mathematica, I found:

So:
$$t\approx1.92985096273712170694334669601\tag6$$
