Question Number 6947 by Tawakalitu. last updated on 03/Aug/16
$${Prove}\:{that}\: \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left\{\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{x}\:−\:{y}}{\left({x}\:+\:{y}\right)^{\mathrm{3}} }\:\:{dy}\right\}\:{dx}\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by Yozzii last updated on 03/Aug/16
$${Q}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−{y}}{\left({x}+{y}\right)^{\mathrm{3}} }{dy} \\ $$$${Let}\:{u}={x}+{y}\Rightarrow{du}={dy} \\ $$$$\Rightarrow{y}={u}−{x} \\ $$$${At}\:{y}=\mathrm{0},{u}={x};\:{at}\:{y}=\mathrm{1},{u}={x}+\mathrm{1} \\ $$$$\therefore{Q}\left({x}\right)=\int_{{x}} ^{{x}+\mathrm{1}} \frac{{x}−\left({u}−{x}\right)}{{u}^{\mathrm{3}} }{du} \\ $$$${Q}\left({x}\right)=\int_{{x}} ^{{x}+\mathrm{1}} {u}^{−\mathrm{3}} \left(\mathrm{2}{x}−{u}\right){du}=\int_{{x}} ^{{x}+\mathrm{1}} \left(\mathrm{2}{xu}^{−\mathrm{3}} −{u}^{−\mathrm{2}} \right){du} \\ $$$${Q}\left({x}\right)=\frac{\mathrm{2}{xu}^{−\mathrm{2}} }{−\mathrm{2}}−\frac{{u}^{−\mathrm{1}} }{−\mathrm{1}}\mid_{{x}} ^{{x}+\mathrm{1}} \\ $$$${Q}\left({x}\right)=\frac{\mathrm{1}}{{u}}−\frac{{x}}{{u}^{\mathrm{2}} }\mid_{{x}} ^{{x}+\mathrm{1}} \\ $$$${Q}\left({x}\right)=\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{{x}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}}+\frac{{x}}{{x}^{\mathrm{2}} } \\ $$$${Q}\left({x}\right)=\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{2}\left({x}+\mathrm{1}−\mathrm{1}\right)}{\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${Q}\left({x}\right)=\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$\therefore{I}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left\{\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{x}−{y}}{\left({x}+{y}\right)^{\mathrm{3}} }{dy}\right\}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {Q}\left({x}\right){dx} \\ $$$${I}=\left[{ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\left({x}+\mathrm{1}\right)^{\mathrm{2}} \mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${I}=\left[{ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×{ln}\mid{x}+\mathrm{1}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${I}=\frac{−\mathrm{1}}{{x}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{1}} =\frac{−\mathrm{1}}{\mathrm{2}}−\left(\frac{−\mathrm{1}}{\mathrm{1}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−{y}}{\left({x}+{y}\right)^{\mathrm{3}} }{dydx}=\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 03/Aug/16
$${You}\:{are}\:{making}\:{my}\:{day}…\:{Thanks}\:{so}\:{much}..\:{am}\:{really}\:{happy} \\ $$