a. ∫ [((1−e^x )/(1+e^x ))]^(1/2) dx=? b. ∫ ((lnx)/(√(1+x)))=? c. ∫


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Question Number 57819 by behi83417@gmail.com last updated on 12/Apr/19

$a . \int [\frac{1 - e^{x}}{1 + e^{x}}]^{\frac{1}{2}} dx = ?$ $b . \int \frac{lnx}{\sqrt{1 + x}} = ?$ $c . \underset{\sqrt{e}}{\int^{e}} \sin (lnx) dx = ?$
Commented by Abdo msup. last updated on 13/Apr/19

$b) l e t I = \int \frac{l n (x)}{\sqrt{1 + x}} d x b y p a r t s u^{^{'}} = \frac{1}{\sqrt{1 + x}} a n d v = l n (x)$ $\Rightarrow I = 2 \sqrt{1 + x} l n (x) - \int \frac{2 \sqrt{1 + x}}{x} d x$ $= 2 l n (x) \sqrt{1 + x} - 2 \int \frac{\sqrt{1 + x}}{x} d x b u t$ $\int \frac{\sqrt{1 + x}}{x} d x =_{\sqrt{1 + x} = t} \int \frac{t}{t^{2} - 1} (2 t) d t$ $= 2 \int \frac{t^{2}}{t^{2} - 1} d t = 2 \int \frac{t^{2} - 1 + 1}{t^{2} - 1} d t$ $= 2 t + 2 \int \frac{d t}{t^{2} - 1} = 2 t + \int (\frac{1}{t - 1} - \frac{1}{t + 1}) d t$ $= 2 t + l n ∣ \frac{t - 1}{t + 1} ∣ + c$ $= 2 \sqrt{1 + x} + l n ∣ \frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1} ∣ + c \Rightarrow$ $I = 2 \sqrt{1 + x} l n (x) - 4 \sqrt{1 + x} - 2 l n ∣ \frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1} ∣ + c .$
Commented by behi83417@gmail.com last updated on 13/Apr/19

$t h a n k y o u v e r y m u c h p r o p h . A b d o .$
Commented by Abdo msup. last updated on 13/Apr/19

$y o u a r e w e l c o m e .$
Commented by Abdo msup. last updated on 13/Apr/19
$let I =∫(√((1−e^x )/(1+e^x )))dx changement e^x =t give I =∫ (√((1−t)/(1+t)))(dt/t) now we use ch.(√((1−t)/(1+t)))=u ⇒ ((1−t)/(1+t)) =u^2 ⇒1−t =u^2 +u^2 t ⇒1−u^2 =(1+u^2 )t ⇒ t =((1−u^2 )/(1+u^2 )) ⇒dt =((−2u(1+u^2 )−2u(1−u^2 ))/((1+u^2 )^2 )) =((−4u)/((1+u^2 )^2 ))du ⇒I =∫ u ((1+u^2 )/(1−u^2 )) ((−4u)/((1+u^2 )^2 )) du =−4 ∫ (u^2 /((1−u^2 )(1+u^2 ))) du =−2 ∫u^2 { (1/(1−u^2 )) −(1/(1+u^2 ))}du =2 ∫ (u^2 /(u^2 −1)) du +2∫ (u^2 /(1+u^2 ))du =2∫ ((u^2 −1+1)/(u^2 −1)) du +2 ∫((1+u^2 −1)/(1+u^2 )) du =2u +∫ (2/(u^2 −1)) du +2u −2 ∫ (du/(1+u^2 )) =4u +∫ ((1/(u−1)) −(1/(u+1)))du −2 ∫ (du/(1+u^2 )) =4u +ln∣((u−1)/(u+1))∣−2arctan(u)+c =4(√((1−t)/(1+t))) +ln∣(((√((1−t)/(1+t)))−1)/((√((1−t)/(1+t)))+1))∣−2arctan((√((1−t)/(1+t))))+c I=4(√((1−e^x )/(1+e^x ))) +ln∣(((√((1−e^x )/(1+e^x )))−1)/((√((1−e^x )/(1+e^x )))+1))∣−2arctan((√((1−e^x )/(1+e^x ))))+c$
$l e t I = \int \sqrt{\frac{1 - e^{x}}{1 + e^{x}}} d x c h a n g e m e n t e^{x} = t g i v e$ $I = \int \sqrt{\frac{1 - t}{1 + t}} \frac{d t}{t} n o w w e u s e c h . \sqrt{\frac{1 - t}{1 + t}} = u \Rightarrow$ $\frac{1 - t}{1 + t} = u^{2} \Rightarrow 1 - t = u^{2} + u^{2} t \Rightarrow 1 - u^{2} = (1 + u^{2}) t \Rightarrow$ $t = \frac{1 - u^{2}}{1 + u^{2}} \Rightarrow d t = \frac{- 2 u (1 + u^{2}) - 2 u (1 - u^{2})}{{(1 + u^{2})}^{2}}$ $= \frac{- 4 u}{{(1 + u^{2})}^{2}} d u \Rightarrow I = \int u \frac{1 + u^{2}}{1 - u^{2}} \frac{- 4 u}{{(1 + u^{2})}^{2}} d u$ $= - 4 \int \frac{u^{2}}{(1 - u^{2}) (1 + u^{2})} d u$ $= - 2 \int u^{2} {\frac{1}{1 - u^{2}} - \frac{1}{1 + u^{2}}} d u$ $= 2 \int \frac{u^{2}}{u^{2} - 1} d u + 2 \int \frac{u^{2}}{1 + u^{2}} d u$ $= 2 \int \frac{u^{2} - 1 + 1}{u^{2} - 1} d u + 2 \int \frac{1 + u^{2} - 1}{1 + u^{2}} d u$ $= 2 u + \int \frac{2}{u^{2} - 1} d u + 2 u - 2 \int \frac{d u}{1 + u^{2}}$ $= 4 u + \int (\frac{1}{u - 1} - \frac{1}{u + 1}) d u - 2 \int \frac{d u}{1 + u^{2}}$ $= 4 u + l n ∣ \frac{u - 1}{u + 1} ∣ - 2 a r c t a n (u) + c$ $= 4 \sqrt{\frac{1 - t}{1 + t}} + l n ∣ \frac{\sqrt{\frac{1 - t}{1 + t}} - 1}{\sqrt{\frac{1 - t}{1 + t}} + 1} ∣ - 2 a r c t a n (\sqrt{\frac{1 - t}{1 + t}}) + c$ $I = 4 \sqrt{\frac{1 - e^{x}}{1 + e^{x}}} + l n ∣ \frac{\sqrt{\frac{1 - e^{x}}{1 + e^{x}}} - 1}{\sqrt{\frac{1 - e^{x}}{1 + e^{x}}} + 1} ∣ - 2 a r c t a n (\sqrt{\frac{1 - e^{x}}{1 + e^{x}}}) + c$
Commented by behi83417@gmail.com last updated on 13/Apr/19

$t h a n k s i n a d v a m c e d e a r p r o p h . A b d o .$
Commented by peter frank last updated on 13/Apr/19

$n i c e w o r k$
Commented by maxmathsup by imad last updated on 19/Apr/19

$y o u a r e w e l c o m e$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19
	$c)∫_(√e) ^e sin(lnx)dx t=lnx→x=e^t dx=e^t dt ∫_(1/2) ^1 sint×e^t dt now use formula ∫e^(ax) sinbxdx=((e^(ax) (asinbx−bcosbx))/(a^2 +b^2 )) ∣((e^t (sint−cost))/(1^2 +1^2 ))∣_(1/2) ^1 (1/2){e(sin1−cos1)−(√e) (sin(1/2)−cos(1/2))}$
	$c) \int_{\sqrt{e}}^{e} s i n (l n x) d x$ $t = l n x \to x = e^{t} d x = e^{t} d t$ $\int_{\frac{1}{2}}^{1} s i n t \times e^{t} d t$ $n o w u s e f o r m u l a$ $\int e^{a x} s i n b x d x = \frac{e^{a x} (a s i n b x - b c o s b x)}{a^{2} + b^{2}}$ $∣ \frac{e^{t} (s i n t - c o s t)}{1^{2} + 1^{2}} ∣_{\frac{1}{2}}^{1}$ $\frac{1}{2} {e (s i n 1 - c o s 1) - \sqrt{e} (s i n \frac{1}{2} - c o s \frac{1}{2})}$
	Commented by behi83417@gmail.com last updated on 13/Apr/19

	$t h a n k y o u s o m u c h s i r t a n m a y .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

	$m o s t s e l c o m e . . . p l s c h e c k a n s w e r o f a)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

	$a) \int \frac{1 - e^{x}}{\sqrt{1 - e^{2 x}}} d x$ $\int \frac{d x}{\sqrt{1 - e^{2 x}}} - \int \frac{e^{x} d x}{\sqrt{1 - e^{2 x}}}$ $\int \frac{e^{- x} d x}{\sqrt{e^{- 2 x} - 1}} - \int \frac{e^{x} d x}{\sqrt{1 - e^{2 x}}}$ $= (- 1) \int \frac{d (e^{- x})}{\sqrt{{(e^{- x})}^{2} - 1}} - \int \frac{d (e^{x})}{\sqrt{1 - {(e^{x})}^{2}}}$ $n o w u s e f o r m u l a$ $\int \frac{d x}{\sqrt{x^{2} - a^{2}}} = l n (x + \sqrt{x^{2} - a^{2}})$ $\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = {s i n}^{- 1} (\frac{x}{a})$ $= (- 1) l n (e^{- x} + \sqrt{e^{- 2 x} - 1}) - {s i n}^{- 1} (\frac{e^{x}}{1}) + c$
	Commented by behi83417@gmail.com last updated on 13/Apr/19

	$y o u r a n s w e r i s r i g h t s i r t a n m a y .$ $t h a n k s i n a d v a n c e .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

	$m o s t w e l c o m e s i r$
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