The provided image contains a mathematical approximation for $h(1.06)$, which involves the function $h(x) = \frac{\ln(x-1)}{(x+1)^2}$. Here's a breakdown of the explanation:

1. Expression for $h(1.06)$: The function $h(x) = \frac{\ln(x-1)}{(x+1)^2}$ is evaluated at $x = 1.06$. This means: $h(1.06) = \frac{\ln(1.06 - 1)}{(1.06 + 1)^2} = \frac{\ln(0.06)}{(2.06)^2}.$

2. Approximation of $h(1.06)$ using linearization: To simplify the calculation, the linear approximation $h(1.06) \approx L(1.06)$ is used. This involves finding the linearization $L(x)$ of $h(x)$ around a nearby point.

3. Linear approximation formula: The linearization $L(x)$ around a point $a$ is: $L(x) = h(a) + h'(a)(x-a).$ Here, the point of expansion $a$ seems to be chosen as $1.06$, but the context suggests the use of properties or approximations directly.

4. Approximation result: The value $h(1.06)$ is approximated as: $h(1.06) \approx L(1.06) = -\frac{1}{4}(1.06 - 1).$

   Simplifying $1.06 - 1 = 0.06$, this becomes: $h(1.06) \approx -\frac{1}{4}(0.06) = -0.015.$

5. Interpretation of $-0.015$: The boxed value $-0.015$ is the linear approximation of $h(1.06)$. It avoids directly computing $\ln(0.06)$ and the denominator $(2.06)^2$, which might involve more complex calculations.

Would you like me to provide more details or explore a specific aspect of this approximation? Here are some related questions to deepen understanding:

1. How is $h'(x)$ derived, and why is $h'(1.06) = -\frac{1}{4}$?
2. What is the role of linear approximation in simplifying function evaluations?
3. Can this approach be extended to approximate $h(x)$ for other nearby values of $x$?
4. What happens if the expansion point $a$ is chosen to be a value other than $1.06$?
5. How accurate is this linear approximation compared to the exact calculation?

Tip: Linear approximations work best when $x$ is close to the expansion point $a$, so choosing $a$ wisely ensures better accuracy.