Question Number 122919 by bemath last updated on 20/Nov/20
Answered by Dwaipayan Shikari last updated on 20/Nov/20
![∫_0 ^(1/2) ((xe^(2x) )/((1+2x)^2 ))dx =(1/4)∫_0 ^1 ((ue^u )/((1+u)^2 ))du =(1/4)∫_0 ^1 (e^u /((u+1)))−(e^u /((u+1)^2 )) =(1/4)[(e^u /(u+1))]_0 ^1 =(1/4)((e/2)−1)=(1/8)(e−2)](https://www.tinkutara.com/question/Q122920.png)
$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{xe}^{\mathrm{2}{x}} }{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ue}^{{u}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{{u}} }{\left({u}+\mathrm{1}\right)}−\frac{{e}^{{u}} }{\left({u}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{e}^{{u}} }{{u}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{e}}{\mathrm{2}}−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left({e}−\mathrm{2}\right) \\ $$
Answered by liberty last updated on 20/Nov/20
![let 1+2x = u ⇒x=((u−1)/2) ∧ dx = (1/2)du φ = ∫ (((((u−1)/2))e^(u−1) )/u^2 ) (1/2)du = (1/4)∫ (((u−1)e^(u−1) )/u^2 )du φ = (1/(4e)) ∫ (((u−1)e^u )/u^2 ) du = (1/(4e))[∫ ((u.e^u −e^u )/u^2 ) du] φ = (1/(4e))∫ d((e^u /u)) = (1/(4e))((e^u /u))=(e^(u−1) /(4u)) put border { ((u=2)),((u=1)) :} give φ = (1/4) [(e^(u−1) /u) ]_1 ^2 = (1/4)((e/2)−1).](https://www.tinkutara.com/question/Q122921.png)
$${let}\:\mathrm{1}+\mathrm{2}{x}\:=\:{u}\:\Rightarrow{x}=\frac{{u}−\mathrm{1}}{\mathrm{2}}\:\wedge\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{du} \\ $$$$\phi\:=\:\int\:\frac{\left(\frac{{u}−\mathrm{1}}{\mathrm{2}}\right){e}^{{u}−\mathrm{1}} }{{u}^{\mathrm{2}} }\:\frac{\mathrm{1}}{\mathrm{2}}{du}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{\left({u}−\mathrm{1}\right){e}^{{u}−\mathrm{1}} }{{u}^{\mathrm{2}} }{du} \\ $$$$\phi\:=\:\frac{\mathrm{1}}{\mathrm{4}{e}}\:\int\:\frac{\left({u}−\mathrm{1}\right){e}^{{u}} }{{u}^{\mathrm{2}} }\:{du}\:=\:\frac{\mathrm{1}}{\mathrm{4}{e}}\left[\int\:\frac{{u}.{e}^{{u}} −{e}^{{u}} }{{u}^{\mathrm{2}} }\:{du}\right] \\ $$$$\phi\:=\:\frac{\mathrm{1}}{\mathrm{4}{e}}\int\:{d}\left(\frac{{e}^{{u}} }{{u}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}{e}}\left(\frac{{e}^{{u}} }{{u}}\right)=\frac{{e}^{{u}−\mathrm{1}} }{\mathrm{4}{u}} \\ $$$${put}\:{border}\:\begin{cases}{{u}=\mathrm{2}}\\{{u}=\mathrm{1}}\end{cases} \\ $$$${give}\:\phi\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\left[\frac{{e}^{{u}−\mathrm{1}} }{{u}}\:\right]_{\mathrm{1}} ^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{e}}{\mathrm{2}}−\mathrm{1}\right).\: \\ $$