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Question Number 122919 by bemath last updated on 20/Nov/20

	Answered by Dwaipayan Shikari last updated on 20/Nov/20

	$\int_{0}^{\frac{1}{2}} \frac{{x e}^{2 x}}{{(1 + 2 x)}^{2}} d x$ $= \frac{1}{4} \int_{0}^{1} \frac{{u e}^{u}}{{(1 + u)}^{2}} d u$ $= \frac{1}{4} \int_{0}^{1} \frac{e^{u}}{(u + 1)} - \frac{e^{u}}{{(u + 1)}^{2}}$ $= \frac{1}{4} {[\frac{e^{u}}{u + 1}]}_{0}^{1} = \frac{1}{4} (\frac{e}{2} - 1) = \frac{1}{8} (e - 2)$
	Answered by liberty last updated on 20/Nov/20
	$let 1+2x = u ⇒x=((u−1)/2) ∧ dx = (1/2)du φ = ∫ (((((u−1)/2))e^(u−1) )/u^2 ) (1/2)du = (1/4)∫ (((u−1)e^(u−1) )/u^2 )du φ = (1/(4e)) ∫ (((u−1)e^u )/u^2 ) du = (1/(4e))[∫ ((u.e^u −e^u )/u^2 ) du] φ = (1/(4e))∫ d((e^u /u)) = (1/(4e))((e^u /u))=(e^(u−1) /(4u)) put border { ((u=2)),((u=1)) :} give φ = (1/4) [(e^(u−1) /u) ]_1 ^2 = (1/4)((e/2)−1).$
	$l e t 1 + 2 x = u \Rightarrow x = \frac{u - 1}{2} \land d x = \frac{1}{2} d u$ $ϕ = \int \frac{(\frac{u - 1}{2}) e^{u - 1}}{u^{2}} \frac{1}{2} d u = \frac{1}{4} \int \frac{(u - 1) e^{u - 1}}{u^{2}} d u$ $ϕ = \frac{1}{4 e} \int \frac{(u - 1) e^{u}}{u^{2}} d u = \frac{1}{4 e} [\int \frac{u . e^{u} - e^{u}}{u^{2}} d u]$ $ϕ = \frac{1}{4 e} \int d (\frac{e^{u}}{u}) = \frac{1}{4 e} (\frac{e^{u}}{u}) = \frac{e^{u - 1}}{4 u}$ $p u t b o r d e r {\begin{cases} u = 2 \\ u = 1 \end{cases}$ $g i v e ϕ = \frac{1}{4} {[\frac{e^{u - 1}}{u}]}_{1}^{2} = \frac{1}{4} (\frac{e}{2} - 1) .$
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