<s>[INST] Integral of $\sqrt{2+t^2}$ How can I solve the integral of 
$$ \int \sqrt{2+t^2}\,dt$$
I obtained it by calculating the length of the arc $a(t)=(t\cos t,t \sin t,t)$ with $t \in [0,2\pi]$. 
I try with substitution with the function $\sinh$ but i don't manage to resolve it.
 [/INST] One way is to integrate by parts:
$$
  \int_0^{2\pi} \underbrace{\sqrt{t^2 + 2}}_u \, \underbrace{dt}_{dv}
    = \underbrace{\sqrt{t^2 + 2}}_u \cdot \underbrace{t}_v \bigg|_0^{2\pi}
        - \int_0^{2\pi} \underbrace{t}_v \cdot \underbrace{\frac t{\sqrt{t^2+2}} \, dt}_{du}
$$
Continuing without the labels:
\begin{align*}
  I &= \int_0^{2\pi} \sqrt{t^2 + 2} \, dt\\[0.3cm]
    &= t\sqrt{t^2 + 2} \bigg|_0^{2\pi} - \int_0^{2\pi} \frac{t^2}{\sqrt{t^2+2}} \, dt\\[0.3cm]
    &= 2\pi \sqrt{4\pi^2+2} - \int_0^{2\pi} \frac{t^2 \color{red}{+2-2}}{\sqrt{t^2+2}} \, dt\\[0.3cm]
    &= 2\pi \sqrt{4\pi^2+2} - \left[\int_0^{2\pi} \frac{t^2 + 2}{\sqrt{t^2+2}} \, dt - \int_0^{2\pi} \frac 2{\sqrt{t^2+2}} \, dt\right]\\[0.3cm]
    &= 2\pi \sqrt{4\pi^2+2} - \underbrace{\int_0^{2\pi} \sqrt{t^2+2} \, dt}_{\text{this is $I$}} + \int_0^{2\pi} \frac 2{\sqrt{t^2+2}} \, dt\\[0.3cm]
    &= 2\pi \sqrt{4\pi^2+2} - I + 2 \sinh^{-1}\left(\frac t{\sqrt2}\right)\bigg|_0^{2\pi}\\[0.3cm]
    &= 2\pi \sqrt{4\pi^2+2} - I + 2\sinh^{-1}(\pi\sqrt2)
\end{align*}
Therefore
$$ I = 2\pi \sqrt{4\pi^2+2} - I + 2\sinh^{-1}(\pi\sqrt2). $$
Solve for $I$ to get
$$ I = \pi \sqrt{4\pi^2+2} + \sinh^{-1}(\pi\sqrt2). $$
Admittedly this does require knowledge of that inverse hyperbolic sine integral (although there are other ways to evaluate that one), which may make this method not that helpful, but here it is for the sake of variety.
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