Let $\gamma(t)=te^{it}$ for $0 \leq t \leq \pi$.
I want to evaluate $$ \int_\gamma z(1+\left|z\right|^2)^{-1/2}\,\left|dz\right|. $$ Substituting and simplifying we have:
$$\int_{0}^{\pi}te^{it}(1+t^2)^{-1/2}\left|tie^{it}+e^{it}\right|\,dt=\int_{0}^{\pi}te^{it}(1+t^2)^{-1/2}\left(t^2+1\right)\,dt=\int_{0}^{\pi}te^{it}(1+t^2)^{1/2}\,dt$$
Is there some sort of trick that I can use to approach the last integral? I don't have the advantage of being able to use residues or Cauchy's various theorems, so I'm looking for a calculus trick so I can solve this by hand.
$$ dz=(\mathrm{e}^{it}+it\mathrm{e}^{it})\,dt, $$ and thus $$ |dz|=\sqrt{1+t^2}\,dt. $$ tHus you have $$ \int_\gamma \frac{z\,dz}{\sqrt{1+|z|^2}}=\int_0^\pi \frac{t\mathrm{e}^{it}\sqrt{1+t^2}\,dt}{\sqrt{1+t^2}}=\int_0^\pi t\mathrm{e}^{it}\,dt=(-it+1)\mathrm{e}^{it}\big|_0^\pi=\cdots. $$