Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 12

Determine the function $\displaystyle h(x) = \ln \left( x + \sqrt{x^2 - 1} \right)$

$
\begin{equation}
\begin{aligned}
h'(x) &= \frac{d}{dx} \ln \left( x+ \sqrt{x^2-1} \right)\\
\\
h'(x) &= \frac{1}{x+\sqrt{x^2-1}} \cdot \frac{d}{dx} \left( x + \sqrt{x^2+1} \right)\\
\\
h'(x) &= \frac{1}{x+\sqrt{x^2-1}} \cdot \frac{d}{dx} \left[ x + \left( x^2 -1 \right)^{\frac{1}{2}}\right]\\
\\
h'(x) &= \frac{1}{x+\sqrt{x^2-1}} \left[ 1 + \frac{1}{2} \left( x^2 - 1 \right)^{\frac{-1}{2}} \cdot \frac{d}{dx} \left( x^2 - 1 \right) \right]\\
\\
h'(x) &= \frac{1}{x+\sqrt{x^2-1}} \left[ 1 + \frac{\cancel{2}x}{\cancel{2}(x^2 -1)^\frac{1}{2}} \right]\\
\\
h'(x) &= \frac{1}{x+\sqrt{x^2-1}} \left[ 1 + \frac{x}{(x^2-1)^\frac{1}{2}} \right]\\
\\
h'(x) &= \frac{1}{x+\sqrt{x^2-1}} \left[ \frac{(x^2-1)^{\frac{1}{2}}+x}{(x^2-1)^{\frac{1}{2}}} \right]\\
\\
h'(x) &= \frac{(x^2-1)^{\frac{1}{2}}+x}{x(x^2-1)^{\frac{1}{2}}+x^2-1} \qquad \text{ or } \qquad h'(x) = \frac{\sqrt{x^2-1}+x}{x\sqrt{x^2-1}+x^2-1}
\end{aligned}
\end{equation}
$