Determining the best possible substitution for an integrand What substitution is best used to calculate  $$\int \frac{1}{1 + \sqrt{x^2 -1}}dx$$

@Pp. suggests an Euler substitution of the form
\begin{align*}
x  &= \frac{t^2+1}{2t} = \frac{t}{2} + \frac{1}{2t} \\
dx &= \left(\frac{1}{2} - \frac{1}{2t^2}\right)\,dt = \frac{t^2-1}{2t^2}\,dt 
\end{align*}

By this method you gain being able to solve all integrals of this type: Additions, multiplications and divisions of the square root of a quadratic polynomial and $x$.

Then
\begin{align}
x^2 - 1 &= \frac{t^4+2t^2+1}{4t^2} - \frac{4t^2}{4t^2} \\
        &= \frac{t^4-2t^2+1}{4t^2} = \left(\frac{t^2-1}{2t}\right)^2
\end{align}
So the integral becomes
\begin{align*}
\int\frac{1}{1+\sqrt{x^2-1}}\,dx
&= \int\frac{1}{1+\dfrac{t^2-1}{2t}} \left(\frac{t^2-1}{2t^2}\right)\,dt\\
&= \int\frac{t^2-1}{t(t^2+2t-1)}\,dt
\end{align*}

The result of the Euler substitution is the integral of a rational function, which can always be solved by using their partial fraction decomposition.

In this case the denominator splits into linear factors.
$$
\frac{t^2-1}{t(t^2+2t-1)} = \frac{A}{t} + \frac{B}{t+1-\sqrt{2}} + \frac{C}{t+1+\sqrt{2}}
$$
