∫ ((xe^x )/( (√(1+e^x )))) dx


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Question Number 116385 by bemath last updated on 03/Oct/20

$\int \frac{{xe}^{x}}{\sqrt{1 + e^{x}}} dx$
	Answered by MJS_new last updated on 03/Oct/20

	$\int \frac{e^{x} x}{\sqrt{e^{x} + 1}} d x =$ $[t = \sqrt{e^{x} + 1} \to d x = \frac{2 \sqrt{e^{x} + 1}}{e^{x}}]$ $= 2 \int \ln (t^{2} - 1) dt = 2 \int \ln (t - 1) d t + 2 \int \ln (t + 1) d t =$ $= 2 ((t - 1) \ln (t - 1) - t) + 2 ((t + 1) \ln (t + 1) - t) =$ $= 2 t \ln (t^{2} - 1) + 2 \ln \frac{t + 1}{t - 1} - 4 t =$ $= 2 (x - 2) \sqrt{e^{x} + 1} + 2 \ln (e^{x} + 2 + 2 \sqrt{e^{x} + 1}) - 2 x + C$
	Answered by Dwaipayan Shikari last updated on 03/Oct/20

	$\int \frac{{xe}^{x}}{\sqrt{1 + e^{x}}} dx 1 + e^{x} = t^{2} \Rightarrow e^{x} = 2 t \frac{dt}{dx}$ $\int \frac{x 2 tdt}{t} = 2 \int xdt = 2 \int \log (t^{2} - 1) dt$ $2 tlog (t^{2} - 1) - \int \frac{{4 t}^{2}}{t^{2} - 1} dt$ $2 tlog (t^{2} - 1) - 4 t + 2 \log (\frac{t + 1}{t - 1}) + C$ $2 x \sqrt{e^{x} + 1} - 4 \sqrt{e^{x} + 1} + 2 \log (\frac{\sqrt{e^{x} + 1} + 1}{\sqrt{e^{x} + 1} - 1}) + C$ $2 (\sqrt{e^{x} + 1} (x - 2) + \log (\frac{\sqrt{e^{x} + 1} + 1}{\sqrt{e^{x} + 1} - 1}) + C_{1})$
	Commented by MJS_new last updated on 03/Oct/20

	$- \int \frac{4 t^{2}}{t^{2} - 1} d t = - 4 t + 2 \ln \frac{t + 1}{t - 1}$ $beside this minor error your result is the$ $same as mine because$ $\frac{\sqrt{e^{x} + 1} + 1}{\sqrt{e^{x} + 1} - 1} = \frac{e^{x} + 2 + 2 \sqrt{e^{x} + 1}}{e^{x}}$ $\Rightarrow$ $2 \ln \frac{\sqrt{e^{x} + 1} + 1}{\sqrt{e^{x} + 1} - 1} = - 2 x + 2 \ln (e^{x} + 2 + 2 \sqrt{e^{x} + 1})$
	Commented by Dwaipayan Shikari last updated on 03/Oct/20

	$Thanking you for correction . Yes i have noticed that result is$ $same$
	Answered by mathmax by abdo last updated on 03/Oct/20
	$I =∫ ((x e^x )/(√(1+e^x )))dx we do thechangement (√(1+e^x ))=t ⇒1+e^x =t^2 ⇒ e^x =t^2 −1 ⇒x =ln(t^2 −1) ⇒ I =∫ ((ln(t^2 −1)(t^2 −1))/t)×((2tdt)/(t^2 −1)) =2 ∫ln(t^2 −1)dt =_(by parts) 2{ t ln(t^2 −1)−∫ t.((2t)/(t^2 −1))dt} =2t ln(t^2 −1)−4 ∫ ((t^2 −1+1)/(t^2 −1))dt =2tln(t^2 −1)−4t −4 ∫ (dt/(t^2 −1)) =2tln(t^2 −1)−4t−2 ∫((1/(t−1))−(1/(t+1)))dt =2t ln(t^2 −1)−4t −2ln∣((t−1)/(t+1))∣ +c ⇒I =2x(√(1+e^x ))−4(√(1+e^x )) −2ln∣(((√(1+e^x ))−1)/((√(1+e^x ))+1))∣ +c$
	$I = \int \frac{x e^{x}}{\sqrt{1 + e^{x}}} dx we do thechangement \sqrt{1 + e^{x}} = t \Rightarrow 1 + e^{x} = t^{2} \Rightarrow$ $e^{x} = t^{2} - 1 \Rightarrow x = \ln (t^{2} - 1) \Rightarrow$ $I = \int \frac{\ln (t^{2} - 1) (t^{2} - 1)}{t} \times \frac{2 tdt}{t^{2} - 1} = 2 \int \ln (t^{2} - 1) dt$ $=_{by parts} 2 {t \ln (t^{2} - 1) - \int t . \frac{2 t}{t^{2} - 1} dt}$ $= 2 t \ln (t^{2} - 1) - 4 \int \frac{t^{2} - 1 + 1}{t^{2} - 1} dt$ $= 2 tln (t^{2} - 1) - 4 t - 4 \int \frac{dt}{t^{2} - 1}$ $= 2 tln (t^{2} - 1) - 4 t - 2 \int (\frac{1}{t - 1} - \frac{1}{t + 1}) dt$ $= 2 t \ln (t^{2} - 1) - 4 t - 2 \ln ∣ \frac{t - 1}{t + 1} ∣ + c$ $\Rightarrow I = 2 x \sqrt{1 + e^{x}} - 4 \sqrt{1 + e^{x}} - 2 \ln ∣ \frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1} ∣ + c$
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