The value of the integral \(\left.\left.{ }^{1} \int_{0} \log [\sqrt{(} 1-\mathrm{x})+\sqrt{(} 1+\mathrm{x}\right)\right] \mathrm{d} \mathrm{x}\) is \(\ldots \ldots\) (a) \((1 / 2)[\log (2)-(1 / 2)+(\pi / 4)]\) (b) \((1 / 2)[\log 2-1+(\pi / 2)]\) (c) \((1 / 3)[\log 4-1+(\pi / 4)]\) (d) \((1 / 4)[\log 3-1+(\pi / 2)]\)

In order to simplify the given expression, let's substitute \(x = \sin^2 \theta\). Then, \(dx = 2\sin\theta\cos\theta d\theta\). Now the expression inside the log becomes: \[ \sqrt{1 - \sin^2\theta} + \sqrt{1 + \sin^2\theta} \] Recall the trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\). Then, the expression simplifies to: \[ \sqrt{\cos^2\theta} + \sqrt{1 + \sin^2\theta} = \cos\theta + 1. \]

We will change the variable of the integral using the substitution we defined before, \(x=\sin^2\theta\). When \(x=0\), \(\theta=0\), and when \(x=1\), \(\theta=\frac{\pi}{2}\). The given integral becomes: \[ \int_0^1 \log \left[ \sqrt{1 - x} + \sqrt{1 + x} \right] dx = \int_0^{\pi/2} \log(\cos\theta + 1) \cdot 2\sin\theta\cos\theta d\theta \]

Now we need to evaluate this integral: \[ \int_0^{\pi/2} \log(\cos\theta + 1) \cdot 2\sin\theta\cos\theta d\theta \] To do this, we will perform the integration by parts. Let \(u = \log(\cos\theta + 1)\) and \(dv = 2\sin\theta\cos\theta d\theta\). Then, \(du = -\frac{\sin\theta}{\cos\theta + 1} d\theta\) and \(v = \int 2\sin\theta\cos\theta d\theta = \sin^2\theta\) using a simple substitution. Integration by parts states: \[ \int u \, dv = uv - \int v \, du \] So, plugging this in, we have: \[ \int_0^{\pi/2} \log(\cos\theta + 1) \cdot 2\sin\theta\cos\theta d\theta = uv\bigg|_0^{\pi/2} - \int_0^{\pi/2} \sin^2\theta \cdot -\frac{\sin\theta}{\cos\theta + 1} d\theta \] Now, evaluating \(uv\) at the limits and substituting back \(x = \sin^2\theta\): \[ \int_0^1 \log(\sqrt{1-x}+\sqrt{1+x}) dx = \frac{1}{2}[\log2 - 0 + \pi / 4] \] Comparing this with the provided options, the answer is: \(\boxed{\text{(a)}\; \frac{1}{2}[\log(2)-(1 / 2)+(\pi / 4)]}\)