To solve the given surface integral, we follow these steps:

Problem Overview

We need to evaluate the surface integral: $\iint_\sigma \frac{1}{(1+x+z)^2} \, dS,$ where $\sigma$ is the portion of the plane $x + y + z = 1$ in the first octant bounded by the coordinate planes $x = 0$, $y = 0$, and $z = 0$.

Step 1: Parametrize the Plane

The equation of the plane is $x + y + z = 1$. Solving for $z$, we have: $z = 1 - x - y.$ The plane in the first octant is bounded by the coordinate planes and the plane $x + y + z = 1$. Thus, the bounds are: $0 \leq x \leq 1, \quad 0 \leq y \leq 1 - x.$

Step 2: Compute $dS$

The surface element $dS$ is given by: $dS = \sqrt{\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 + 1} \, dx \, dy.$ For $z = 1 - x - y$: $\frac{\partial z}{\partial x} = -1, \quad \frac{\partial z}{\partial y} = -1.$ Thus: $dS = \sqrt{(-1)^2 + (-1)^2 + 1} \, dx \, dy = \sqrt{1 + 1 + 1} \, dx \, dy = \sqrt{3} \, dx \, dy.$

Step 3: Setup the Surface Integral

Substitute $dS = \sqrt{3} \, dx \, dy$ and $z = 1 - x - y$ into the integral: $\iint_\sigma \frac{1}{(1 + x + z)^2} \, dS = \iint_R \frac{1}{(1 + x + (1 - x - y))^2} \sqrt{3} \, dx \, dy,$ where $R$ is the region in the $xy$-plane defined by: $0 \leq x \leq 1, \quad 0 \leq y \leq 1 - x.$

Simplify the integrand: $1 + x + z = 1 + x + (1 - x - y) = 2 - y.$ Thus, the integral becomes: $\iint_R \frac{\sqrt{3}}{(2 - y)^2} \, dx \, dy.$

Step 4: Evaluate the Integral

Factor out $\sqrt{3}$: $\sqrt{3} \int_0^1 \int_0^{1-x} \frac{1}{(2 - y)^2} \, dy \, dx.$ First, evaluate the inner integral with respect to $y$: $\int_0^{1-x} \frac{1}{(2 - y)^2} \, dy.$ Let $u = 2 - y$, so $du = -dy$. When $y = 0$, $u = 2$. When $y = 1-x$, $u = 1+x$. The integral becomes: $-\int_2^{1+x} \frac{1}{u^2} \, du = \left[\frac{1}{u}\right]_2^{1+x} = \frac{1}{2} - \frac{1}{1+x}.$

Thus, the integral becomes: $\sqrt{3} \int_0^1 \left(\frac{1}{2} - \frac{1}{1+x}\right) \, dx.$

Step 5: Evaluate the Outer Integral

Split into two parts: $\sqrt{3} \left[\int_0^1 \frac{1}{2} \, dx - \int_0^1 \frac{1}{1+x} \, dx\right].$

1. First term: $\int_0^1 \frac{1}{2} \, dx = \frac{1}{2} \int_0^1 1 \, dx = \frac{1}{2} [x]_0^1 = \frac{1}{2}.$

2. Second term: $\int_0^1 \frac{1}{1+x} \, dx = \ln|1+x|_0^1 = \ln(2) - \ln(1) = \ln(2).$

Combine the results: $\sqrt{3} \left(\frac{1}{2} - \ln(2)\right).$

Final Answer

$\boxed{\sqrt{3} \left(\frac{1}{2} - \ln(2)\right).}$

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