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 $$\cosh^2y-\sinh^2y=1$$
Hint: recall that
$$ \int \frac{1}{\sqrt{1+w^2}} dw = \sinh^{-1}(w) +C $$
where $\sinh^{-1}$ indicates the inverse of the hyperbolic sine function. Consider the substitution $w=tv$ and proceed.