Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D \frac{dx\,dy}{(x+y+1)^2}, $$ where $0<x<1$ and $0<y<1$.

Step 1 · Set up the integral

$$ I = \int_0^1\int_0^1 \frac{dy\,dx}{(x+y+1)^2}. $$

Step 2 · Integrate with respect to $y$

Treating $x$ as constant, let $u=x+y+1$, $du=dy$:

$$ \int_0^1 \frac{dy}{(x+y+1)^2} = \left[-\frac{1}{x+y+1}\right]_0^1 = -\frac{1}{x+2} + \frac{1}{x+1} = \frac{1}{x+1}-\frac{1}{x+2}. $$

Step 3 · Integrate with respect to $x$

$$ I = \int_0^1 \left(\frac{1}{x+1}-\frac{1}{x+2}\right)dx = \Big[\ln(x+1)-\ln(x+2)\Big]_0^1. $$

Step 4 · Evaluate at the endpoints

At $x=1$: $\ln2-\ln3$. At $x=0$: $\ln1-\ln2=-\ln2$. So:

$$ I = (\ln2-\ln3) - (-\ln2) = 2\ln2-\ln3 = \ln4-\ln3 = \ln\!\left(\frac{4}{3}\right). $$

Step 5 · Numerical check

$$ \ln\!\left(\frac43\right) \approx \ln(1.3333) \approx 0.2877. $$

Since $\frac{1}{(x+y+1)^2}$ ranges from $\frac{1}{9}$ (at $x=y=1$) to $1$ (at $x=y=0$) on the unit square, and the square has area $1$, the estimation theorem gives $\frac19\leqslant I\leqslant1$; the exact value $\approx0.288$ lies comfortably inside this range. ✓

Step 6 · Symmetry check via the other order

Since $(x+y+1)^{-2}$ is symmetric in $x$ and $y$, integrating with respect to $x$ first instead gives the identical computation with $x$ and $y$ swapped, so the same value $\ln\!\left(\frac43\right)$ results either way.

Additional Notes

• Although $\frac{1}{(x+y+1)^2}$ is not a product of a function of $x$ alone and a function of $y$ alone, it is still elementary to integrate in either variable because it depends only on the sum $x+y$.