Question Number 32484 by Eng.Firas last updated on 25/Mar/18
$$ \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+{y}\right)}{\left({x}+{y}\right)}\:{dx}\:{dy} \\ $$
Commented by abdo imad last updated on 26/Mar/18
![I =∫_1 ^2 ( ∫_0 ^1 ((ln(x+y))/(x+y))dx)dy let put A= ∫_0 ^1 ((ln(x+y))/(x+y))dx ch. x+y =t give A = ∫_y ^(1+y) ((lnt)/t) dt =[(1/2)( lnt)^2 ]_y ^(1+y) =(1/2) ( (ln(1+y)^2 −(lny)^2 ) ⇒ I = (1/2) ∫_1 ^2 (ln(1+y))^2 dy − (1/2) ∫_1 ^2 ((lny))^2 dy but ch. lny =t give ∫_1 ^2 (lny)^2 dy = ∫_0 ^(ln(2)) t^2 e^t dt = [t^2 e^t ]_0 ^(ln(2)) −∫_0 ^(ln(2)) 2t e^t dt = 2(ln2)^2 −2 ( [t e^t ]_0 ^(ln(2)) −∫_0 ^(ln(2)) e^t dt) =2(ln2)^2 −2 (2ln2 −1)= 2(ln2)^2 −4ln(2) +1 also ch. ln(1+y) =t give ∫_1 ^2 (ln(1+y))^2 dy = ∫_(ln2) ^(ln(3)) t^2 e^t dt and we get the value of this intevral by the same method.....](https://www.tinkutara.com/question/Q32488.png)
$${I}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\left(\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({x}+{y}\right)}{{x}+{y}}{dx}\right){dy}\:\:{let}\:{put}\:{A}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({x}+{y}\right)}{{x}+{y}}{dx} \\ $$$${ch}.\:{x}+{y}\:={t}\:{give} \\ $$$${A}\:=\:\int_{{y}} ^{\mathrm{1}+{y}} \:\:\frac{{lnt}}{{t}}\:{dt}\:\:=\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\:{lnt}\right)^{\mathrm{2}} \right]_{{y}} ^{\mathrm{1}+{y}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\left(\:\left({ln}\left(\mathrm{1}+{y}\right)^{\mathrm{2}} \:−\left({lny}\right)^{\mathrm{2}} \right)\:\Rightarrow\right. \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left({ln}\left(\mathrm{1}+{y}\right)\right)^{\mathrm{2}} {dy}\:\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left(\left({lny}\right)\right)^{\mathrm{2}} {dy}\:\:{but} \\ $$$${ch}.\:{lny}\:={t}\:{give}\: \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\left({lny}\right)^{\mathrm{2}} \:{dy}\:=\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}\right)} \:{t}^{\mathrm{2}} \:{e}^{{t}} \:{dt}\:=\:\left[{t}^{\mathrm{2}} {e}^{{t}} \right]_{\mathrm{0}} ^{{ln}\left(\mathrm{2}\right)} \:−\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}\right)} \:\mathrm{2}{t}\:{e}^{{t}} {dt} \\ $$$$=\:\mathrm{2}\left({ln}\mathrm{2}\right)^{\mathrm{2}} \:−\mathrm{2}\:\left(\:\left[{t}\:{e}^{{t}} \right]_{\mathrm{0}} ^{{ln}\left(\mathrm{2}\right)} \:−\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}\right)} {e}^{{t}} {dt}\right) \\ $$$$=\mathrm{2}\left({ln}\mathrm{2}\right)^{\mathrm{2}} \:\:−\mathrm{2}\:\left(\mathrm{2}{ln}\mathrm{2}\:−\mathrm{1}\right)=\:\mathrm{2}\left({ln}\mathrm{2}\right)^{\mathrm{2}} \:−\mathrm{4}{ln}\left(\mathrm{2}\right)\:+\mathrm{1}\:{also} \\ $$$${ch}.\:{ln}\left(\mathrm{1}+{y}\right)\:={t}\:{give} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\left({ln}\left(\mathrm{1}+{y}\right)\right)^{\mathrm{2}} \:{dy}\:=\:\int_{{ln}\mathrm{2}} ^{{ln}\left(\mathrm{3}\right)} \:{t}^{\mathrm{2}} \:{e}^{{t}} {dt}\:\:{and}\:{we}\:{get}\:{the}\:{value}\:{of} \\ $$$${this}\:{intevral}\:{by}\:{the}\:{same}\:{method}….. \\ $$
Commented by Eng.Firas last updated on 26/Mar/18
$${Thank}\:{you} \\ $$