Definite integration by substitution problem with the substitution $x=e^u−e^{−u}$ 
I'm stuck at the following integration problem:
a) Given that $x = e^u - e^{-u}$, find an expression for $e^u$ in terms of $x$.
b) Use the substitution $x = e^u - e^{-u}$ to find the exact value of $\int_0^1 \frac {1} {\sqrt {4+x^2}}\, dx$.

My textbook says that the solutions are:
a) $e^u=\frac {x+\sqrt{x^2+4}}{2}$
b) $\ln\frac {1+\sqrt5}{2}$
I have tried some algebraic manipulation in task a) but this did not bring me far. The textbook solutions looks pretty much like the result of the quadratic formula but I do not know if this is a coincidence or not.
I did not understand b) either since I lack the understanding of task a). I might be able to solve it because you could use the substitution $u =\frac x 2 $ because this nicely leads to $\int_0^1 \frac 1 {\sqrt{u^2+1}}\,du $ and then I one could trig-substitution but I cannot solve it with the given substitution.
Could somebody tell me I can solve or even approach the problem?

Multiplying by $e^u$, you get
$$
x = e^u-e^{-u}\Leftrightarrow x e^u = (e^u)^2-1\Leftrightarrow (e^u)^2-x e^u-1=0
$$
solving the quadratic equation, you conclude that
$$
e^u = \frac{x \pm \sqrt{x^2+4}}{2}
$$
but the minus sign does not yield any solution and you finally arrive at
$$
e^u = \frac{x + \sqrt{x^2+4}}{2}.
$$
Regarding the integral,
\begin{align*}
\int_0^1 \frac{1}{\sqrt{x^2+4}}dx = \int_0^{\log \frac{1+\sqrt{5}}{2}} (e^u+e^{-u})\frac{1}{e^u+e^{-u}} du = \log \frac{1+\sqrt{5}}{2}
\end{align*}

Note:
$$
e^u = \frac{x+\sqrt{x^2+4}}{2} \Rightarrow \sqrt{x^2+4} = 2e^u -x = 2e^u-(e^u-e^{-u}) = e^u+e^{-u}
$$
