Question Number 30572 by abdo imad last updated on 23/Feb/18
![find I=∫∫_([3,4]×[1,2]) ((dxdy)/((x+y)^2 )) .](https://www.tinkutara.com/question/Q30572.png)
$${find}\:\:{I}=\int\int_{\left[\mathrm{3},\mathrm{4}\right]×\left[\mathrm{1},\mathrm{2}\right]} \:\:\frac{{dxdy}}{\left({x}+{y}\right)^{\mathrm{2}} }\:. \\ $$
Commented by prof Abdo imad last updated on 24/Feb/18
![I= ∫_1 ^2 (∫_3 ^4 (dx/((x+y)^2 )))dy but we have ∫_3 ^4 (dx/((x+y)^2 ))=[ −(1/(x+y))]_(x=3) ^(x=4) = (1/(y+3)) −(1/(y+4)) ⇒ I= ∫_1 ^2 ( (1/(y+3)) −(1/(y+4)))dy =[ln∣((y+3)/(y+4))∣]_1 ^2 =ln((5/6)) −ln((4/5)) .](https://www.tinkutara.com/question/Q30718.png)
$${I}=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left(\int_{\mathrm{3}} ^{\mathrm{4}} \:\:\:\frac{{dx}}{\left({x}+{y}\right)^{\mathrm{2}} }\right){dy}\:{but}\:{we}\:{have} \\ $$$$\int_{\mathrm{3}} ^{\mathrm{4}} \:\:\:\frac{{dx}}{\left({x}+{y}\right)^{\mathrm{2}} }=\left[\:−\frac{\mathrm{1}}{{x}+{y}}\right]_{{x}=\mathrm{3}} ^{{x}=\mathrm{4}} =\:\frac{\mathrm{1}}{{y}+\mathrm{3}}\:−\frac{\mathrm{1}}{{y}+\mathrm{4}}\:\Rightarrow \\ $$$${I}=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left(\:\frac{\mathrm{1}}{{y}+\mathrm{3}}\:−\frac{\mathrm{1}}{{y}+\mathrm{4}}\right){dy}\:=\left[{ln}\mid\frac{{y}+\mathrm{3}}{{y}+\mathrm{4}}\mid\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$={ln}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\:−{ln}\left(\frac{\mathrm{4}}{\mathrm{5}}\right)\:. \\ $$