Determining the best possible substitution for an integrand What substitution is best used to calculate  $$\int \frac{1}{1 + \sqrt{x^2 -1}}dx$$
 A: The quadratic form suggests $x = \sec\theta$.  But rationalizing first gives
$$
\frac{1}{1 + \sqrt{x^2-1}} = \frac{\sqrt{x^2-1}-1}{x^2} 
$$
Now substituting $x=\sec\theta$ yields (after some algebra)
$$
\int\frac{\sqrt{x^2-1}-1}{x^2}\,dx = \int(\sec\theta - \cos\theta - \sin\theta)\,d\theta
$$
A: @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.
$$
\frac{t^2-1}{t(t^2+2t-1)} = \frac{A}{t} + \frac{B}{t+1-\sqrt{2}} + \frac{C}{t+1+\sqrt{2}}
$$
A: The obvious substitution when dealing with expressions of the form $\sqrt{x^2\pm a^2}$ is either $x=a\sinh t$ $($for $+)$ or $x=a\cosh t$ $($for $-)$. Then use $\cosh^2t-\sinh^2t=1,~$ along with $\sinh't=\cosh t~$ and $~\cosh't=\sinh t$.
