Length of a curve given by $x=e^tcost$, $y=e^tsint$, $z=t$ 
Let $C$ be the curve $x=e^t\cos(t)$, $y=e^t\sin(t)$, $z=t$ between $t=0$ and $t=2\pi$. I want to find the length of the curve.

First we write the vector $r$ as $r(t) = e^t\cos(t)\cdot \hat{i} +e^t\sin(t)\cdot \hat{j} + t\cdot \hat{k}$. The length of it is equal to 
$$\displaystyle\int_0^{2\pi}|dr/dt|dt=\displaystyle\int_0^{2\pi}\sqrt{2e^{2t}+1}dt$$
I am setting $v^2 = 2e^{2t}+1 $ so I get $2e^{2t}dt=vdv$ and my integral becomes 
$$\displaystyle\int\cfrac{v^2}{v^2-1}$$
After calculations, I get a wrong answer. How can I calculate the given integral? 
 A: Given
$$r(t) = e^t\cos(t) \vec{i} +e^t \sin(t)\vec{j} + t\vec{z}$$
$$v(t) = (e^t\cos t-e^t\sin t)\vec{i} +(e^t \sin t+ e^t\cos t )\vec{j} + 1\vec{z}$$
$$|v|^2=2e^{2t}+1$$
and want to calculate 
$$\int_0^{2\pi}\sqrt{2e^{2t}+1}dt$$
So let $2e^{2t}=\sinh^2(u)$, then $4e^{2t}dt=2\sinh(u)\cosh(u) du$ 
or $2\sinh^2(u)dt=2\sinh(u)\cosh(u) du$
ie $dt=\frac{\cosh u}{\sinh u }du$
The integral is after using $1+\sinh^2=\cosh^2$
$$\int \coth(u)\cosh(u) du$$
Integration by parts: $f=\coth,\, df=-\operatorname{csch}^2$, $dg=\cosh, \, g=\sinh$
$$\coth(u)\sinh(u)+\int \frac{1}{\sinh(u)}du$$
$$=\cosh(u)+\ln(\tanh(u/2))$$
$$=\sqrt{1+2e^{2t}}\Big|_0^{2\pi}+\ln\left(\frac{\sqrt{2}e^t}{1+\sqrt{1+2e^{2t}}}\right)\Big|_0^{2\pi}$$
$$=\sqrt{1+2e^{4\pi}}+\ln\left(\frac{\sqrt{2}e^{2\pi}}{1+\sqrt{1+2e^{4\pi}}}\right)-\sqrt{3}-\ln\left(\frac{\sqrt{2}}{1+\sqrt{3}}\right)$$
which I guess simplifies slightly to
$$\boxed{=\sqrt{1+2e^{4\pi}}+2\pi-\ln\left(1+\sqrt{1+2e^{4\pi}}\right)-\sqrt{3}+\ln\left(1+\sqrt{3}\right)}$$
Edit: Might be worth showing that:
$$\int \frac{du}{\sinh(u)}=\int \frac{2e^u\, du}{e^{2u}-1}=\int \frac{2 dv}{v^2-1}=\ln\left(\frac{v-1}{v+1}\right)=\ln\left(\frac{e^u-1}{e^u+1}\right)=\ln(\tanh(u/2))$$
and that
$$\tanh(u/2)=\frac{\sinh(u)}{1+\cosh(u)}$$
and from the above $1+2e^{2t}=1+\sinh^2(u)=\cosh^2(u)$
Edit 2: The other approach is just to do it the way you were
$$\int\sqrt{2e^{2t}+1}dt=\int\frac{v^2}{v^2-1}dv=\int 1+\frac{1}{v^2-1}dv$$
$$=v+\frac{1}{2}\int \frac{1}{v-1}-\frac{1}{v+1}dv=v+\frac{1}{2}\ln\left(\frac{v-1}{v+1}\right)$$
So
$$\int\sqrt{2e^{2t}+1}dt=\sqrt{1+2e^{2t}}\Big|_0^{2\pi}+\frac{1}{2}\ln\left(\frac{\sqrt{1+2e^{2t}}-1}{\sqrt{1+2e^{2t}}+1}\right)\Big|_0^{2\pi}$$
The first expression is the same as the above, the second you can see is the same because
\begin{align}
\frac{1}{2}\ln\left(\frac{\sqrt{1+2e^{2t}}-1}{\sqrt{1+2e^{2t}}+1}\right)&=\frac{1}{2}\ln\left(\frac{\sqrt{1+2e^{2t}}-1}{\sqrt{1+2e^{2t}}+1}\cdot \frac{\sqrt{1+2e^{2t}+1}}{\sqrt{1+2e^{2t}}+1}\right)\\
&=\frac{1}{2}\ln\left(\frac{2e^{2t}}{(\sqrt{1+2e^{2t}}+1)^2}\right)\\
&=\frac{1}{2}\ln\left(\left[\frac{\sqrt{2}e^{t}}{\sqrt{1+2e^{2t}}+1}\right]^2\right)\\
&=\ln\left(\frac{\sqrt{2}e^{t}}{\sqrt{1+2e^{2t}}+1}\right)
\end{align}
A: $$L=\int_0^{2\pi}\sqrt{2e^{2t}+1}dt$$
$$t\to \frac{\ln(u-1/2)}{2}$$
$$L=\int_{3/2}^{e^{4\pi}+1/2}\frac{\sqrt{2u}}{2u-1}du$$
$$u\to v^2$$
$$L=\int_{\sqrt{3/2}}^{\sqrt{e^{4\pi}+1/2}}\frac{\sqrt{2u}}{2u-1}du$$
$$L=\int_{\sqrt{3/2}}^{\sqrt{e^{4\pi}+1/2}}\frac{2v^2\sqrt{2}}{2v^2-1}dv$$
$$L=\sqrt{2}\int_{\sqrt{3/2}}^{\sqrt{e^{4\pi}+1/2}}\frac{2v^2}{2v^2-1}dv$$
$$L=\sqrt{2}\int_{\sqrt{3/2}}^{\sqrt{e^{4\pi}+1/2}}\bigg(1+\frac{1}{2v^2-1}\bigg)dv$$
$$L=\sqrt{2}\int_{\sqrt{3/2}}^{\sqrt{e^{4\pi}+1/2}}\bigg(1+\frac{1}{2v^2-1}\bigg)dv$$
$$L=\sqrt{2}\int_{\sqrt{3/2}}^{\sqrt{e^{4\pi}+1/2}}\bigg(1+\frac{1}{2\sqrt2 v-2}-\frac{1}{2\sqrt2 v+2}\bigg)dv$$
$$L=\sqrt{2}\int_{\sqrt{3/2}}^{\sqrt{e^{4\pi}+1/2}}\bigg(v+\frac{1}{2\sqrt2 v-2}-\frac{1}{2\sqrt2 v+2}\bigg)dv$$
Yuck, I'll let you take it from here.
The reason you got the wrong answer is that you dropped some constants along the way, like the $\sqrt 2$.
