Question Number 78706 by abdomathmax last updated on 20/Jan/20
![calculate ∫∫_([0,1]^2 ) ((dxdy)/((x+y+1)^2 ))](https://www.tinkutara.com/question/Q78706.png)
$${calculate}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\frac{{dxdy}}{\left({x}+{y}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Answered by mind is power last updated on 20/Jan/20
![=∫_0 ^1 −[(1/(x+y+1))]_0 ^1 dx =∫_0 ^1 −(1/(x+2))−(1/(x+1))dx =−ln(3)+ln(2)−ln(2)=ln((1/3))](https://www.tinkutara.com/question/Q78736.png)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} −\left[\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}\mathrm{dx} \\ $$$$=−\mathrm{ln}\left(\mathrm{3}\right)+\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{2}\right)=\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$