# Evaluate integral if R=[0,1]x[0,1]

1. Jul 24, 2014

### jonroberts74

1. The problem statement, all variables and given/known data
evaluate integral if R=[0,1]x[0,1]

$$\iint_R ln[(x+1)(y+1)]dA$$

3. The attempt at a solution

$$\int_{0}^{1}\int_{0}^{1} ln[(x+1)(y+1)] dydx$$

by parts

$$u=ln[(x+1)(y+1)]; du = \frac{1}{(x+1)(y+1)}dy;dv=dy;v=y$$

$$\int_{0}^{1} \Bigg[\frac{y}{(x+1)(y+1)} - \frac{1}{x+1}\int_{0}^{1} 1 - \frac{1}{y+1}dy\Bigg]dx$$

$$\int_{0}^{1} \Bigg[\frac{1}{2(x+1)} - \frac{1}{x+1} - \frac{ln(2)}{(x+1)}\Bigg]dx$$

$$\frac{1}{2}ln(2) - ln(2) - ln^2(2)$$

not sure what I did wrong but I know thats the incorrect answer

2. Jul 24, 2014

### SammyS

Staff Emeritus
An error was made when you stated $$u=ln[(x+1)(y+1)];\ \ du = \frac{1}{(x+1)(y+1)}dy\ .$$

Treat (x+1) as a constant (with respect to u). Then: Either use the chain rule, or write $\ \ln[(x+1)(y+1)]\$ as $\ \ln(x+1)+\ln(y+1)\$

3. Jul 24, 2014

### jonroberts74

so I could do

$$\int_{0}^{1}ln(x+1)dx+\int_{0}^{1}ln(y+1)dy$$

forgot about the log properties for a second

or even

$$2\int_{0}^{1}ln(x+1)dx$$ or the y counterpart

$$2[(x+1)ln(x+1)-(x+1)]\Bigg|_{0}^{1} = 4ln(2)-2 = ln(16)-2$$

Last edited: Jul 24, 2014
4. Jul 24, 2014

### SammyS

Staff Emeritus
You can't generally do that sort of split for a double integral (or any iterated integral for that matter).

However, the following is true.

$\displaystyle \int_{0}^{1}\int_{0}^{1} ln[(x+1)(y+1)]\, dydx=\int_{0}^{1}\int_{0}^{1} ln(x+1)\, dydx+\int_{0}^{1}\int_{0}^{1} ln(y+1)\, dydx$