let f(x) = ∫_(−1) ^x (e^t /(√(1−e^t )))dt with x<0 1) calculate f(x) 2) find ∫


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Question Number 40152 by maxmathsup by imad last updated on 16/Jul/18

$l e t f (x) = \int_{- 1}^{x} \frac{e^{t}}{\sqrt{1 - e^{t}}} d t w i t h x < 0$ $1) c a l c u l a t e f (x)$ $2) f i n d \int_{- 1}^{0} \frac{e^{t}}{\sqrt{1 - e^{t}}} d t$
Commented by maxmathsup by imad last updated on 16/Jul/18
$changement e^t =u give t=ln(u) ⇒ f(x)= ∫_e^(−1) ^e^x (u/(√(1−u))) (du/u) = ∫_e^(−1) ^e^x (du/(√(1−u))) =[−2(√(1−u))]_e^(−1) ^e^x f(x)=−2{(√(1−e^x )) −(√(1−e^(−1) ))} 2) ∫_(−1) ^0 (e^t /(√(1−e^t ))) dt =lim_(x→0) f(x)= 2(√(1−e^(−1) ))$
$c h a n g e m e n t e^{t} = u g i v e t = l n (u) \Rightarrow$ $f (x) = \int_{e^{- 1}}^{e^{x}} \frac{u}{\sqrt{1 - u}} \frac{d u}{u} = \int_{e^{- 1}}^{e^{x}} \frac{d u}{\sqrt{1 - u}} = {[- 2 \sqrt{1 - u}]}_{e^{- 1}}^{e^{x}}$ $f (x) = - 2 {\sqrt{1 - e^{x}} - \sqrt{1 - e^{- 1}}}$ $2) \int_{- 1}^{0} \frac{e^{t}}{\sqrt{1 - e^{t}}} d t = {l i m}_{x \to 0} f (x) = 2 \sqrt{1 - e^{- 1}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jul/18
	$∫_(−1) ^x (e^t /(√(1−e^t )))dt =−1∫_(−1) ^x ((d(1−e^t ))/(√(1−e^t ))) =−1×∣((√(1−e^t ))/(1/2))∣_(−1) ^x =−2{(√(1−e^x )) −(√(1−e^(−1) )) } ∫_(−1) ^0 (e^t /((√(1−e^t )) ))dt =−2{(√(1−e^0 )) −(√(1−e_ ^(−1) })) =2(√(1−e^(−1) )) }$
	$\int_{- 1}^{x} \frac{e^{t}}{\sqrt{1 - e^{t}}} d t$ $= - 1 \int_{- 1}^{x} \frac{d (1 - e^{t})}{\sqrt{1 - e^{t}}}$ $= - 1 \times ∣ \frac{\sqrt{1 - e^{t}}}{\frac{1}{2}} ∣_{- 1}^{x}$ $= - 2 {\sqrt{1 - e^{x}} - \sqrt{1 - e^{- 1}}}$ $\int_{- 1}^{0} \frac{e^{t}}{\sqrt{1 - e^{t}}} d t$ $= - 2 {\sqrt{1 - e^{0}} - \sqrt{1 - e_{}^{- 1}}}$ $= 2 \sqrt{1 - e^{- 1}}}$
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