Let $ a,\ b$ are postive constant numbers.

(1) Differentiate $ \ln (x\plus{}\sqrt{x^2\plus{}a})\ (x>0).$

(2) For $ a\equal{}\frac{4b^2}{(e\minus{}e^{\minus{}1})^2}$, evaluate $ \int_0^b \frac{1}{\sqrt{x^2\plus{}a}}\ dx.$
1. Differentiate \( \ln (x + \sqrt{x^2 + a}) \) for \( x > 0 \).

   To differentiate \( \ln (x + \sqrt{x^2 + a}) \), we use the chain rule. Let \( u = x + \sqrt{x^2 + a} \). Then, we have:
   \[
   \frac{d}{dx} \ln u = \frac{1}{u} \cdot \frac{du}{dx}
   \]
   Now, we need to find \( \frac{du}{dx} \):
   \[
   u = x + \sqrt{x^2 + a}
   \]
   Differentiating \( u \) with respect to \( x \):
   \[
   \frac{du}{dx} = 1 + \frac{d}{dx} \sqrt{x^2 + a}
   \]
   Using the chain rule on \( \sqrt{x^2 + a} \):
   \[
   \frac{d}{dx} \sqrt{x^2 + a} = \frac{1}{2\sqrt{x^2 + a}} \cdot 2x = \frac{x}{\sqrt{x^2 + a}}
   \]
   Therefore:
   \[
   \frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 + a}}
   \]
   Substituting back into the derivative of \( \ln u \):
   \[
   \frac{d}{dx} \ln (x + \sqrt{x^2 + a}) = \frac{1}{x + \sqrt{x^2 + a}} \left( 1 + \frac{x}{\sqrt{x^2 + a}} \right)
   \]
   Simplifying the expression inside the parentheses:
   \[
   1 + \frac{x}{\sqrt{x^2 + a}} = \frac{\sqrt{x^2 + a} + x}{\sqrt{x^2 + a}}
   \]
   Thus:
   \[
   \frac{d}{dx} \ln (x + \sqrt{x^2 + a}) = \frac{1}{x + \sqrt{x^2 + a}} \cdot \frac{\sqrt{x^2 + a} + x}{\sqrt{x^2 + a}} = \frac{\sqrt{x^2 + a} + x}{(x + \sqrt{x^2 + a}) \sqrt{x^2 + a}}
   \]
   Notice that \( x + \sqrt{x^2 + a} \) in the numerator and denominator cancel out:
   \[
   \frac{d}{dx} \ln (x + \sqrt{x^2 + a}) = \frac{1}{\sqrt{x^2 + a}}
   \]

2. Evaluate \( \int_0^b \frac{1}{\sqrt{x^2 + a}} \, dx \) for \( a = \frac{4b^2}{(e - e^{-1})^2} \).

   To evaluate the integral, we use the substitution \( x = \sqrt{a} \tan \theta \). Then:
   \[
   dx = \sqrt{a} \sec^2 \theta \, d\theta
   \]
   The integral becomes:
   \[
   \int_0^b \frac{1}{\sqrt{x^2 + a}} \, dx = \int_0^{\theta_b} \frac{1}{\sqrt{a \tan^2 \theta + a}} \cdot \sqrt{a} \sec^2 \theta \, d\theta
   \]
   Simplifying inside the square root:
   \[
   \sqrt{a \tan^2 \theta + a} = \sqrt{a (\tan^2 \theta + 1)} = \sqrt{a \sec^2 \theta} = \sqrt{a} \sec \theta
   \]
   Thus, the integral becomes:
   \[
   \int_0^{\theta_b} \frac{\sqrt{a} \sec^2 \theta}{\sqrt{a} \sec \theta} \, d\theta = \int_0^{\theta_b} \sec \theta \, d\theta
   \]
   The integral of \( \sec \theta \) is:
   \[
   \int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta| + C
   \]
   Evaluating from \( 0 \) to \( \theta_b \), where \( \theta_b \) is such that \( \sqrt{a} \tan \theta_b = b \):
   \[
   \theta_b = \tan^{-1} \left( \frac{b}{\sqrt{a}} \right)
   \]
   Therefore:
   \[
   \int_0^b \frac{1}{\sqrt{x^2 + a}} \, dx = \left. \ln |\sec \theta + \tan \theta| \right|_0^{\theta_b}
   \]
   At \( \theta = 0 \):
   \[
   \sec 0 = 1, \quad \tan 0 = 0 \quad \Rightarrow \quad \ln |1 + 0| = \ln 1 = 0
   \]
   At \( \theta = \theta_b \):
   \[
   \sec \theta_b = \sqrt{1 + \tan^2 \theta_b} = \sqrt{1 + \left( \frac{b}{\sqrt{a}} \right)^2} = \sqrt{\frac{a + b^2}{a}} = \frac{\sqrt{a + b^2}}{\sqrt{a}}
   \]
   \[
   \tan \theta_b = \frac{b}{\sqrt{a}}
   \]
   Therefore:
   \[
   \ln \left| \frac{\sqrt{a + b^2}}{\sqrt{a}} + \frac{b}{\sqrt{a}} \right| = \ln \left| \frac{\sqrt{a + b^2} + b}{\sqrt{a}} \right| = \ln \left( \frac{\sqrt{a + b^2} + b}{\sqrt{a}} \right)
   \]
   Simplifying:
   \[
   \int_0^b \frac{1}{\sqrt{x^2 + a}} \, dx = \ln \left( \frac{\sqrt{a + b^2} + b}{\sqrt{a}} \right)
   \]
   Given \( a = \frac{4b^2}{(e - e^{-1})^2} \):
   \[
   \sqrt{a} = \frac{2b}{e - e^{-1}}
   \]
   Substituting \( a \) into the integral result:
   \[
   \int_0^b \frac{1}{\sqrt{x^2 + \frac{4b^2}{(e - e^{-1})^2}}} \, dx = \ln \left( \frac{\sqrt{\frac{4b^2}{(e - e^{-1})^2} + b^2} + b}{\frac{2b}{e - e^{-1}}} \right)
   \]
   Simplifying inside the logarithm:
   \[
   \sqrt{\frac{4b^2}{(e - e^{-1})^2} + b^2} = \sqrt{b^2 \left( \frac{4}{(e - e^{-1})^2} + 1 \right)} = b \sqrt{\frac{4}{(e - e^{-1})^2} + 1}
   \]
   Therefore:
   \[
   \int_0^b \frac{1}{\sqrt{x^2 + \frac{4b^2}{(e - e^{-1})^2}}} \, dx = \ln \left( \frac{b \sqrt{\frac{4}{(e - e^{-1})^2} + 1} + b}{\frac{2b}{e - e^{-1}}} \right)
   \]
   Simplifying further:
   \[
   = \ln \left( \frac{\sqrt{\frac{4}{(e - e^{-1})^2} + 1} + 1}{\frac{2}{e - e^{-1}}} \right)
   \]
   \[
   = \ln \left( \frac{(e - e^{-1}) (\sqrt{\frac{4}{(e - e^{-1})^2} + 1} + 1)}{2} \right)
   \]

The final answer is \( \boxed{ \ln \left( \frac{(e - e^{-1}) (\sqrt{\frac{4}{(e - e^{-1})^2} + 1} + 1)}{2} \right) } \)