Question: I have to solve this integral but I don't understand which substitution I have to use. I've tried with $\sqrt{v^2+w^2}=t$, $v^2+w^2=t$, adding and subtracting $1$ in the numerator, adding and subtracting $2w$ under square root, but no dice.

Answer: 
we have:
$$I=\int\_0^{\sqrt{1-v^2}}\frac 1{\sqrt{v^2+w^2}}dw=\frac 1v\int\_0^{\sqrt{1-v^2}}\frac 1{\sqrt{1+(w/v)^2}}dw$$
now if we let $x=\frac wv\Rightarrow dw=vdx$ and so:
$$I=\int\_0^{\frac{\sqrt{1-v^2}}{v}}\frac 1{\sqrt{1+x^2}}dx$$
which is now a standard integral that can be computed by letting $x=\sinh(y)$ and remembering that $$\cosh^2y-\sinh^2y=1$$
For convergence, we require $0 < |v| \leq 1$, and then without loss of generatlity we can take $v > 0$. The trick here is to let $w = v \sinh x$, so that
\begin{aligned}
v^2+w^2 &= v^2 (1 + \sinh^2 x) = v^2 \cosh^2 x, \\
dw &= v \cosh x \; dx, \\
x &= \sinh^{-1} \frac{w}{v}
\end{aligned}
Our integral then becomes
$$\int\_0^{\sinh^{-1} \sqrt{v^{-2}-1}} \frac{v \cosh x}{\sqrt{v^2 \cosh^2 x}} \; dx = \sinh^{-1} \sqrt{\frac{1}{v^2}-1}$$
since the integrand simplifies to become trivial.