Evaluate the double integral of 1/(x + y + 2) over the square [0,1] x [0,1].

Step-by-step solution

1. 1

   Integrate in $x$ first, treating $y$ as a constant. The integrand is a reciprocal of a linear function of $x$, so $\int_{0}^{1}\frac{dx}{x+y+2}=\Big[\ln(x+y+2)\Big]_{0}^{1}=\ln(y+3)-\ln(y+2).$ Both arguments stay positive on the square, so no absolute values are needed. Order does not matter here - the integrand is symmetric in $x$ and $y$.

2. 2

   Set up the remaining single integral. $\int_{0}^{1}\!\!\int_{0}^{1}\frac{dx\,dy}{x+y+2}=\int_{0}^{1}\Big[\ln(y+3)-\ln(y+2)\Big]dy$ Each piece is a logarithm of a shifted variable, which integrates by parts (or by the standard formula) to $\int\ln(y+a)\,dy=(y+a)\ln(y+a)-(y+a)+C.$

3. 3

   Evaluate the two log integrals. $\int_{0}^{1}\ln(y+3)\,dy=\Big[(y+3)\ln(y+3)-(y+3)\Big]_{0}^{1}=\left(4\ln 4-4\right)-\left(3\ln 3-3\right)=4\ln 4-3\ln 3-1,$ $\int_{0}^{1}\ln(y+2)\,dy=\Big[(y+2)\ln(y+2)-(y+2)\Big]_{0}^{1}=\left(3\ln 3-3\right)-\left(2\ln 2-2\right)=3\ln 3-2\ln 2-1.$ Numerically these are $1.2493405785$ and $0.9095425049$.

4. 4

   Subtract and collect the logarithms. $\left(4\ln 4-3\ln 3-1\right)-\left(3\ln 3-2\ln 2-1\right)=4\ln 4+2\ln 2-6\ln 3.$ Since $4\ln 4=4(2\ln 2)=8\ln 2$, this becomes $8\ln 2+2\ln 2-6\ln 3=10\ln 2-6\ln 3.$

5. 5

   Write it as a single logarithm and be careful with the algebra. Combining, $10\ln 2-6\ln 3=\ln\frac{2^{10}}{3^{6}}=\ln\frac{1024}{729}=0.3397980736.$ A tempting but wrong rewrite is $2\ln 2+4\ln\tfrac43$, which expands to $10\ln 2-4\ln 3=2.5370$ - about seven times too large, because it loses two factors of $3$ in the denominator.

6. 6

   Check the size against a crude estimate. On the unit square $x+y+2$ ranges from $2$ to $4$, so the integrand lies between $\tfrac14$ and $\tfrac12$ and the average must too. The value $0.3398$ sits comfortably in that band, and a $4000\times 4000$ midpoint grid gives $0.3397980732$, confirming the closed form.