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Question Number 80312 by TawaTawa last updated on 02/Feb/20

Commented by mr W last updated on 02/Feb/20

$w h a t i s {. . .} h e r e ?$
Commented by TawaTawa last updated on 02/Feb/20

$(\frac{x - 1}{x + 1}) sir$
Commented by mr W last updated on 02/Feb/20

$w h e t h e r {x} = (x) o r {x} = x - [x], h a v e$ $y o u e v e r t r i e d t o s o l v e i t b y y o u r s e l f ?$
Commented by TawaTawa last updated on 02/Feb/20

$I did it like normal bracket . So i thought i could confirm the$ $answers . But now it is different definition .$ $I cannot solve that one .$
Commented by mr W last updated on 02/Feb/20
$with {((x − 1)/(x + 1))}=(((x − 1)/(x + 1))) the question makes not much sense. i think it means fraction part, i.e. {x}=x−[x]$
$w i t h {\frac{x - 1}{x + 1}} = (\frac{x - 1}{x + 1}) t h e q u e s t i o n$ $m a k e s n o t m u c h s e n s e . i t h i n k i t$ $m e a n s f r a c t i o n p a r t, i . e .$ ${x} = x - [x]$
Commented by TawaTawa last updated on 02/Feb/20

$Maybe sir . I just guess sir .$ $Since {} has meaning . Help me sir .$
Commented by TawaTawa last updated on 02/Feb/20
$Sir, it means the fractional part.$
$Sir, it means the fractional part .$
Commented by john santu last updated on 02/Feb/20
${((x−1)/(x+1))}= ((x−1)/(x+1))−[((x−1)/(x+1))]=((x−1)/(x+1))−[1−(2/(x+1))] =((x−1)/(x+1))−1= ((x−1−x−1)/(x+1))=((−2)/(x+1)) ∗ Ω= ∫ (((−2)/(x+1))).((xdx)/(1−x)) = = ∫ (( 2xdx)/((x^2 −1))) = ∫ ((d(x^2 −1))/(x^2 −1)) Ω= ln(x^2 −1)+c$
${\frac{x - 1}{x + 1}} = \frac{x - 1}{x + 1} - [\frac{x - 1}{x + 1}] = \frac{x - 1}{x + 1} - [1 - \frac{2}{x + 1}]$ $= \frac{x - 1}{x + 1} - 1 = \frac{x - 1 - x - 1}{x + 1} = \frac{- 2}{x + 1}$ $* Ω = \int (\frac{- 2}{x + 1}) . \frac{x d x}{1 - x} =$ $= \int \frac{2 x d x}{(x^{2} - 1)} = \int \frac{d (x^{2} - 1)}{x^{2} - 1}$ $Ω = l n (x^{2} - 1) + c$
Commented by TawaTawa last updated on 02/Feb/20

$God bless you sir .$
Commented by john santu last updated on 02/Feb/20

$m y a n s i s r i g h t ?$
Commented by TawaTawa last updated on 02/Feb/20

$I {don}^{'} t know sir . Sir mrW can confirm .$
Commented by john santu last updated on 02/Feb/20

$o k m i s s . t h a n k y o u . g o d b l e s s$ $y o u$
Commented by behi83417@gmail.com last updated on 02/Feb/20

$Ω = \int \frac{- 2 x dx}{x^{2} - 1} = - \ln ∣ x^{2} - 1 ∣ + const .$
Commented by mathmax by abdo last updated on 02/Feb/20
$if {((x−1)/(x+1))} means (((x−1)/(x+1))) ⇒Ω=∫_0 ^1 ((x−1)/(x+1))×((xdx)/(1−x)) =−∫_0 ^1 ((xdx)/(x+1)) =−∫_0 ^1 ((x+1−1)/(x+1))dx =−1 +∫_0 ^1 (dx/(x+1)) =−1+ln(2) if {((x−1)/(x+1))} means ((x−1)/(x+1))−[((x−1)/(x+1))] changement x=(1/t) give Ω=−∫_1 ^(+∞) {(((1/t)−1)/((1/t)+1))}×(1/(t(1−(1/t))))(−(dt/t^2 )) =∫_1 ^(+∞) {((1−t)/(1+t))}(dt/(t^2 (t−1))) =∫_1 ^(+∞) {((1−t)/(1+t))−[((1−t)/(1+t))])(dt/(t^2 (t−1))) =−∫_1 ^(+∞) (dt/(t^2 (t+1))) −∫_1 ^(+∞) [((1−t)/(1+t))](dt/(t^2 (t−1))) let decompose f(t)=(1/(t^2 (t+1))) ⇒f(t)=(a/t) +(b/t^2 ) +(c/(t+1)) b=t^2 f(t)∣_(t=0) =1 and c=(t+1)f(t)∣_(t=−1) =1 ⇒ f(t)=(a/t)+(1/(t+1)) +(1/t^2 ) f(1)=(1/2) =a+(1/2) +1 ⇒a=−1 ⇒f(t)=−(1/t)+(1/(t+1))+(1/t^2 ) ⇒ ∫_1 ^(+∞) f(t)dt =[−ln∣t∣+ln∣t+1∣−(1/t)]_1 ^(+∞) =[ln∣((t+1)/t)∣−(1/t)]_1 ^(+∞) =−ln(2)+1 also ∫_1 ^(+∞) [((1−t)/(1+t))](dt/(t^2 (t−1))) =∫_1 ^(+∞) [−((t−1)/(t+1))](dt/(t^2 (t−1))) =∫_1 ^(+∞) [−((t+1−2)/(t+1))](dt/(t^2 (t−1))) =∫_1 ^(+∞) [−1+(2/(t+1))](dt/(t^2 (t−1))) =−∫_1 ^(+∞) (dt/(t^2 (t−1))) +∫_1 ^(+∞) [(2/(t+1))](dt/(t^2 (t−1))) ∫_1 ^(+∞) (dt/(t^2 (t−1))) =_(t−1=u) ∫_0 ^(+∞) (du/(u(u+1)^(2 ) )) this integral is divergent..! also ∫_1 ^(+∞) [(2/(t+1))](dt/(t^2 (t−1))) =_(t+1=u) ∫_2 ^(+∞) [(2/u)](du/((u−1)^2 (u−2))) for u>2 ⇒0<(1/u)<(1/2) ⇒0<(2/u)<1 ⇒[(2/u)]=0 any way Ω diverges...$
$i f {\frac{x - 1}{x + 1}} m e a n s (\frac{x - 1}{x + 1}) \Rightarrow Ω = \int_{0}^{1} \frac{x - 1}{x + 1} \times \frac{x d x}{1 - x}$ $= - \int_{0}^{1} \frac{x d x}{x + 1} = - \int_{0}^{1} \frac{x + 1 - 1}{x + 1} d x = - 1 + \int_{0}^{1} \frac{d x}{x + 1} = - 1 + l n (2)$ $i f {\frac{x - 1}{x + 1}} m e a n s \frac{x - 1}{x + 1} - [\frac{x - 1}{x + 1}] c h a n g e m e n t x = \frac{1}{t} g i v e$ $Ω = - \int_{1}^{+ \infty} {\frac{\frac{1}{t} - 1}{\frac{1}{t} + 1}} \times \frac{1}{t (1 - \frac{1}{t})} (- \frac{d t}{t^{2}})$ $= \int_{1}^{+ \infty} {\frac{1 - t}{1 + t}} \frac{d t}{t^{2} (t - 1)} = \int_{1}^{+ \infty} {\frac{1 - t}{1 + t} - [\frac{1 - t}{1 + t}]) \frac{d t}{t^{2} (t - 1)}$ $= - \int_{1}^{+ \infty} \frac{d t}{t^{2} (t + 1)} - \int_{1}^{+ \infty} [\frac{1 - t}{1 + t}] \frac{d t}{t^{2} (t - 1)} l e t d e c o m p o s e$ $f (t) = \frac{1}{t^{2} (t + 1)} \Rightarrow f (t) = \frac{a}{t} + \frac{b}{t^{2}} + \frac{c}{t + 1}$ $b = t^{2} f (t) ∣_{t = 0} = 1 a n d c = (t + 1) f (t) ∣_{t = - 1} = 1 \Rightarrow$ $f (t) = \frac{a}{t} + \frac{1}{t + 1} + \frac{1}{t^{2}}$ $f (1) = \frac{1}{2} = a + \frac{1}{2} + 1 \Rightarrow a = - 1 \Rightarrow f (t) = - \frac{1}{t} + \frac{1}{t + 1} + \frac{1}{t^{2}} \Rightarrow$ $\int_{1}^{+ \infty} f (t) d t = {[- l n ∣ t ∣ + l n ∣ t + 1 ∣ - \frac{1}{t}]}_{1}^{+ \infty} = {[l n ∣ \frac{t + 1}{t} ∣ - \frac{1}{t}]}_{1}^{+ \infty}$ $= - l n (2) + 1 a l s o$ $\int_{1}^{+ \infty} [\frac{1 - t}{1 + t}] \frac{d t}{t^{2} (t - 1)} = \int_{1}^{+ \infty} [- \frac{t - 1}{t + 1}] \frac{d t}{t^{2} (t - 1)}$ $= \int_{1}^{+ \infty} [- \frac{t + 1 - 2}{t + 1}] \frac{d t}{t^{2} (t - 1)} = \int_{1}^{+ \infty} [- 1 + \frac{2}{t + 1}] \frac{d t}{t^{2} (t - 1)}$ $= - \int_{1}^{+ \infty} \frac{d t}{t^{2} (t - 1)} + \int_{1}^{+ \infty} [\frac{2}{t + 1}] \frac{d t}{t^{2} (t - 1)}$ $\int_{1}^{+ \infty} \frac{d t}{t^{2} (t - 1)} =_{t - 1 = u} \int_{0}^{+ \infty} \frac{d u}{u {(u + 1)}^{2}} t h i s i n t e g r a l i s d i v e r g e n t . .!$ $a l s o \int_{1}^{+ \infty} [\frac{2}{t + 1}] \frac{d t}{t^{2} (t - 1)} =_{t + 1 = u} \int_{2}^{+ \infty} [\frac{2}{u}] \frac{d u}{{(u - 1)}^{2} (u - 2)}$ $f o r u > 2 \Rightarrow 0 < \frac{1}{u} < \frac{1}{2} \Rightarrow 0 < \frac{2}{u} < 1 \Rightarrow [\frac{2}{u}] = 0 a n y w a y Ω d i v e r g e s . . .$
Commented by TawaTawa last updated on 02/Feb/20

$God bless you sir$
Commented by TawaTawa last updated on 02/Feb/20

$I appreciate$
Commented by mathmax by abdo last updated on 04/Feb/20

$y o u a r e w e l c o m e .$
	Answered by mr W last updated on 02/Feb/20
	${x}=x−[x] ((x−1)/(x+1))=1−(2/(x+1)) for 0≤x≤1: −1≤((x−1)/(x+1))≤0 concerning [x] for x<0 there are two different ways for the definition: way 1: [x]=⌊x⌋, i.e. [−0.5]=−1 way 2: [x]=⌈x⌉, i.e. [−0.5]=0 if we take definition way 1: since −1≤((x−1)/(x+1))≤0, {((x−1)/(x+1))}=((x−1)/(x+1))−(−1)=((2x)/(x+1)) ∫_0 ^( 1) {((x−1)/(x+1))}((xdx)/(1−x)) =∫_0 ^( 1) ((2x)/(x−1))×((xdx)/(1−x)) =−∫_0 ^( 1) ((2x^2 )/((x−1)^2 ))dx =−2∫_0 ^( 1) (((x−1+1)^2 )/((x−1)^2 ))d(x−1) =−2∫_(−1) ^( 0) (((u+1)^2 )/u^2 )du =−2∫_(−1) ^( 0) (1+(2/u)+(1/u^2 ))du =−2[u+2ln u−(1/u)]_(−1) ^0 =∞ if we take definition way 2: since −1≤((x−1)/(x+1))≤0, {((x−1)/(x+1))}=((x−1)/(x+1))−0=((x−1)/(x+1)) ∫_0 ^( 1) {((x−1)/(x+1))}((xdx)/(1−x)) =∫_0 ^( 1) ((x−1)/(x+1))×((xdx)/(1−x)) =−∫_0 ^( 1) (x/(x+1))dx =−∫_0 ^( 1) (1−(1/(x+1)))dx =−[x−ln (x+1)]_0 ^1 =ln 2−1$
	${x} = x - [x]$ $\frac{x - 1}{x + 1} = 1 - \frac{2}{x + 1}$ $f o r 0 ⩽ x ⩽ 1 :$ $- 1 ⩽ \frac{x - 1}{x + 1} ⩽ 0$ $c o n c e r n i n g [x] f o r x < 0 t h e r e a r e t w o$ $d i f f e r e n t w a y s f o r t h e d e f i n i t i o n :$ $w a y 1 : [x] = ⌊ x ⌋, i . e . [- 0.5] = - 1$ $w a y 2 : [x] = ⌈ x ⌉, i . e . [- 0.5] = 0$ $i f w e t a k e d e f i n i t i o n w a y 1 :$ $s i n c e - 1 ⩽ \frac{x - 1}{x + 1} ⩽ 0,$ ${\frac{x - 1}{x + 1}} = \frac{x - 1}{x + 1} - (- 1) = \frac{2 x}{x + 1}$ $\int_{0}^{1} {\frac{x - 1}{x + 1}} \frac{x d x}{1 - x}$ $= \int_{0}^{1} \frac{2 x}{x - 1} \times \frac{x d x}{1 - x}$ $= - \int_{0}^{1} \frac{2 x^{2}}{{(x - 1)}^{2}} d x$ $= - 2 \int_{0}^{1} \frac{{(x - 1 + 1)}^{2}}{{(x - 1)}^{2}} d (x - 1)$ $= - 2 \int_{- 1}^{0} \frac{{(u + 1)}^{2}}{u^{2}} d u$ $= - 2 \int_{- 1}^{0} (1 + \frac{2}{u} + \frac{1}{u^{2}}) d u$ $= - 2 {[u + 2 \ln u - \frac{1}{u}]}_{- 1}^{0}$ $= \infty$ $i f w e t a k e d e f i n i t i o n w a y 2 :$ $s i n c e - 1 ⩽ \frac{x - 1}{x + 1} ⩽ 0,$ ${\frac{x - 1}{x + 1}} = \frac{x - 1}{x + 1} - 0 = \frac{x - 1}{x + 1}$ $\int_{0}^{1} {\frac{x - 1}{x + 1}} \frac{x d x}{1 - x}$ $= \int_{0}^{1} \frac{x - 1}{x + 1} \times \frac{x d x}{1 - x}$ $= - \int_{0}^{1} \frac{x}{x + 1} d x$ $= - \int_{0}^{1} (1 - \frac{1}{x + 1}) d x$ $= - {[x - \ln (x + 1)]}_{0}^{1}$ $= \ln 2 - 1$
	Commented by TawaTawa last updated on 02/Feb/20

	$God bless you sir .$
	Commented by TawaTawa last updated on 02/Feb/20

	$I appreciate$
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