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Question Number 45020 by rahul 19 last updated on 07/Oct/18

Commented by maxmathsup by imad last updated on 07/Oct/18
$let decompose F(x) =(1/(x^3 +1)) ⇒F(x)=(1/((x+1)(x^2 −x+1))) =(a/(x+1)) +((bx+c)/(x^2 −x +1)) a =lim_(x→−1) (x+1)F(x) =(1/3) lim_(x→+∞) xF(x) =0 =a+b ⇒b =−(1/3) ⇒ F(x)=(1/(3(x+1))) +((−(1/3)x+c)/(x^2 −x+1)) F(0) =(1/3) +c =1 ⇒F(x) =(2/3) ⇒F(x) =(1/(3(x+1))) +(1/3)((−x+2)/(x^2 −x+1)) ⇒ ∫_0 ^1 F(x)dx =(1/3) ∫_0 ^1 (dx/(x+1)) −(1/3) ∫_0 ^1 ((x−2)/(x^2 −x +1))dx =(1/3)ln(2) −(1/6) ∫_0 ^1 ((2x−1−3)/(x^2 −x +1))dx =((ln(2))/3) −(1/6)[ln∣x^2 −x+1∣]_0 ^1 +(1/2) ∫_0 ^1 (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)t) ((ln(2))/3) +(1/2) (4/3)∫_(−(1/(√3))) ^(1/(√3)) (1/(t^2 +1)) ((√3)/2)dt =((ln(2))/3) +(1/(√3)) { arctan((1/(√3)))+arctan((1/(√3)))} =((ln(2))/3) +(2/(√3)) (π/6) =((ln(2))/3) +(π/(3(√3))) =λ .$
$l e t d e c o m p o s e F (x) = \frac{1}{x^{3} + 1} \Rightarrow F (x) = \frac{1}{(x + 1) (x^{2} - x + 1)}$ $= \frac{a}{x + 1} + \frac{b x + c}{x^{2} - x + 1}$ $a = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{1}{3}$ ${l i m}_{x \to + \infty} x F (x) = 0 = a + b \Rightarrow b = - \frac{1}{3} \Rightarrow F (x) = \frac{1}{3 (x + 1)} + \frac{- \frac{1}{3} x + c}{x^{2} - x + 1}$ $F (0) = \frac{1}{3} + c = 1 \Rightarrow F (x) = \frac{2}{3} \Rightarrow F (x) = \frac{1}{3 (x + 1)} + \frac{1}{3} \frac{- x + 2}{x^{2} - x + 1} \Rightarrow$ $\int_{0}^{1} F (x) d x = \frac{1}{3} \int_{0}^{1} \frac{d x}{x + 1} - \frac{1}{3} \int_{0}^{1} \frac{x - 2}{x^{2} - x + 1} d x$ $= \frac{1}{3} l n (2) - \frac{1}{6} \int_{0}^{1} \frac{2 x - 1 - 3}{x^{2} - x + 1} d x$ $= \frac{l n (2)}{3} - \frac{1}{6} {[l n ∣ x^{2} - x + 1 ∣]}_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}$ $=_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} t} \frac{l n (2)}{3} + \frac{1}{2} \frac{4}{3} \int_{- \frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{1}{t^{2} + 1} \frac{\sqrt{3}}{2} d t = \frac{l n (2)}{3} + \frac{1}{\sqrt{3}} {a r c t a n (\frac{1}{\sqrt{3}}) + a r c t a n (\frac{1}{\sqrt{3}})}$ $= \frac{l n (2)}{3} + \frac{2}{\sqrt{3}} \frac{π}{6} = \frac{l n (2)}{3} + \frac{π}{3 \sqrt{3}} = λ .$
Commented by maxmathsup by imad last updated on 07/Oct/18
$let calculate L let A_n = {Π_(k=1) ^n (1+(k^2 /n^2 ))}^(1/n) ⇒ ln(A_n ) =(1/n) Σ_(k=1) ^n ln(1+(k^2 /n^2 )) is a Reiman sum ⇒ lim_(n→+∞) ln(A_n ) =∫_0 ^1 ln(1+x^2 )dx by parts =[xln(1+x^2 )]_0 ^1 −∫_0 ^1 x ((2x)/(1+x^2 ))dx =ln(2)−2 ∫_0 ^1 ((1+x^2 −1)/(1+x^2 ))dx =ln(2)−2 +2 ∫_0 ^1 (dx/(1+x^2 )) =ln(2)−2 +2 [arctanx]_0 ^1 =ln(2)−2+2 (π/4) =ln(2)−2+(π/2) lim_(n→+∞) A_n = e^(ln(2)−2 +(π/2)) = 2 (e^(π/2) /e^2 ) =L we have ln(L) =ln(2)−2 +(π/2) and 3λ =ln(2) +(π/(√3)) ⇒ 3λ +ln(L) =2ln(2) −2 +((1/2) +(1/(√3)))π so any answer given is correct...$
$l e t c a l c u l a t e L l e t A_{n} = {\prod_{k = 1}^{n} (1 + \frac{k^{2}}{n^{2}})}^{\frac{1}{n}} \Rightarrow$ $l n (A_{n}) = \frac{1}{n} \sum_{k = 1}^{n} l n (1 + \frac{k^{2}}{n^{2}}) i s a R e i m a n s u m \Rightarrow$ ${l i m}_{n \to + \infty} l n (A_{n}) = \int_{0}^{1} l n (1 + x^{2}) d x b y p a r t s$ $= {[x l n (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} x \frac{2 x}{1 + x^{2}} d x = l n (2) - 2 \int_{0}^{1} \frac{1 + x^{2} - 1}{1 + x^{2}} d x$ $= l n (2) - 2 + 2 \int_{0}^{1} \frac{d x}{1 + x^{2}} = l n (2) - 2 + 2 {[a r c t a n x]}_{0}^{1} = l n (2) - 2 + 2 \frac{π}{4}$ $= l n (2) - 2 + \frac{π}{2} {l i m}_{n \to + \infty} A_{n} = e^{l n (2) - 2 + \frac{π}{2}} = 2 \frac{e^{\frac{π}{2}}}{e^{2}} = L$ $w e h a v e l n (L) = l n (2) - 2 + \frac{π}{2} a n d 3 λ = l n (2) + \frac{π}{\sqrt{3}} \Rightarrow$ $3 λ + l n (L) = 2 l n (2) - 2 + (\frac{1}{2} + \frac{1}{\sqrt{3}}) π s o a n y a n s w e r g i v e n i s c o r r e c t . . .$
Commented by maxmathsup by imad last updated on 07/Oct/18

$s o a n y a n s w e r g i v e n i s n o t c o r r e c t . . .$
Commented by rahul 19 last updated on 08/Oct/18
thanks prof Abdo☺️
Commented by maxmathsup by imad last updated on 08/Oct/18

$y o u a r e w e l c o m e s i r .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18
	$λ=∫_0 ^1 (dx/(1+x^3 )) (1/(1+x^3 ))=(a/(1+x))+((bx+c)/(1−x+x^2 )) 1=a(1−x+x^2 )+(1+x)(bx+c) 1=(a−ax+ax^2 )+(bx+c+bx^2 +cx) 1=(a+c)+x(−a+b+c)+x^2 (a+b) a+c=1 −a+b+c=0 a+b=0 −a−a+1−a=0 a=(1/3) b=−(1/3) c=(2/3) λ=∫_0 ^1 {(a/(1+x))+((bx+c)/(1−x+x^2 ))}dx λ=(1/3)∫_0 ^1 (dx/(1+x))+∫_0 ^1 ((((−1)/3)x+(2/3))/(1−x+x^2 ))dx λ=(1/3)∫_0 ^1 (dx/(1+x))+(1/3)∫_0 ^1 ((−x+2)/(1−x+x^2 ))dx =(1/3)∫_0 ^1 (dx/(1+x))−(1/3)×(1/2)∫_0 ^1 ((2x−1−3)/(1−x+x^2 ))dx =(1/3)∫_0 ^1 (dx/(1+x))−(1/6)∫_0 ^1 ((d(1−x+x^2 ))/(1−x+x^2 ))dx+(1/2)∫_0 ^1 (dx/(x^2 −2.x.(1/2)+(1/4)+(3/4))) =(1/3)∫_0 ^1 (dx/(1+x))−(1/6)∫_0 ^1 ((d(1−x+x^2 ))/(1−x+x^2 ))+(1/2)∫_0 ^1 (dx/((x−(1/2))^2 +((((√3) )/2))^2 )) =(1/3)∣ln(1+x)∣_0 ^1 −(1/6)∣ln(1−x+x^2 )∣_0 ^1 +(1/2)×(2/(√3))∣tan^(−1) (((x−(1/2))/(((√3) )/2)))∣_0 ^1 =(1/3)ln2+(1/(√3)){tan^(−1) ((1/(√3)))−tan^(−1) (((−1)/(√3)))} =(1/3)ln2+(1/(√3))×((2π)/6) =(1/3)ln2+(π/(3(√3))) wait to find L$
	$λ = \int_{0}^{1} \frac{d x}{1 + x^{3}}$ $\frac{1}{1 + x^{3}} = \frac{a}{1 + x} + \frac{b x + c}{1 - x + x^{2}}$ $1 = a (1 - x + x^{2}) + (1 + x) (b x + c)$ $1 = (a - a x + {a x}^{2}) + (b x + c + {b x}^{2} + c x)$ $1 = (a + c) + x (- a + b + c) + x^{2} (a + b)$ $a + c = 1$ $- a + b + c = 0$ $a + b = 0$ $- a - a + 1 - a = 0 a = \frac{1}{3} b = - \frac{1}{3} c = \frac{2}{3}$ $λ = \int_{0}^{1} {\frac{a}{1 + x} + \frac{b x + c}{1 - x + x^{2}}} d x$ $λ = \frac{1}{3} \int_{0}^{1} \frac{d x}{1 + x} + \int_{0}^{1} \frac{\frac{- 1}{3} x + \frac{2}{3}}{1 - x + x^{2}} d x$ $λ = \frac{1}{3} \int_{0}^{1} \frac{d x}{1 + x} + \frac{1}{3} \int_{0}^{1} \frac{- x + 2}{1 - x + x^{2}} d x$ $= \frac{1}{3} \int_{0}^{1} \frac{d x}{1 + x} - \frac{1}{3} \times \frac{1}{2} \int_{0}^{1} \frac{2 x - 1 - 3}{1 - x + x^{2}} d x$ $= \frac{1}{3} \int_{0}^{1} \frac{d x}{1 + x} - \frac{1}{6} \int_{0}^{1} \frac{d (1 - x + x^{2})}{1 - x + x^{2}} d x + \frac{1}{2} \int_{0}^{1} \frac{d x}{x^{2} - 2 . x . \frac{1}{2} + \frac{1}{4} + \frac{3}{4}}$ $= \frac{1}{3} \int_{0}^{1} \frac{d x}{1 + x} - \frac{1}{6} \int_{0}^{1} \frac{d (1 - x + x^{2})}{1 - x + x^{2}} + \frac{1}{2} \int_{0}^{1} \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}$ $= \frac{1}{3} ∣ l n (1 + x) ∣_{0}^{1} - \frac{1}{6} ∣ l n (1 - x + x^{2}) ∣_{0}^{1} + \frac{1}{2} \times \frac{2}{\sqrt{3}} ∣ {t a n}^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) ∣_{0}^{1}$ $= \frac{1}{3} l n 2 + \frac{1}{\sqrt{3}} {{t a n}^{- 1} (\frac{1}{\sqrt{3}}) - {t a n}^{- 1} (\frac{- 1}{\sqrt{3}})}$ $= \frac{1}{3} l n 2 + \frac{1}{\sqrt{3}} \times \frac{2 π}{6}$ $= \frac{1}{3} l n 2 + \frac{π}{3 \sqrt{3}}$ $w a i t t o f i n d L$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18
	$L=lim_(n→∞) [Π_(r=1) ^n (((n^2 +r^2 ))/n^2 )]^(1/n) =lim_(n→∞) [{1+((1/n))^2 }{1+((2/n))^2 }{1+((3/n))^2 }...{1+((n/n))^2 }]^(1/n) h=(1/n) lnL=lim_(h→0) h[ln(1+1^2 h^2 )+ln(1+2^2 h^2 )+ln(1+3^2 h^2 )+...+ln(1+n^2 h^2 )] lnL=∫_0 ^1 ln(1+x^2 )dx I_1 =∫ln(1+x^2 )dx =ln(1+x^2 )×x−∫((2x×x)/(1+x^2 ))dx =xln(1+x^2 )−2∫((1+x^2 −1)/(1+x^2 ))dx =xln(1+x^2 )−2∫(1−(1/(1+x^2 )))dx =xln(1+x^2 )−2[x−tan^(−1) x] lnL=∣xln(1+x^2 )−2x+2tan^(−1) x∣_0 ^1 lnL=[{ln2−2+2tan^(−1) 1}] lnL=ln2−2+2×(π/4) lnL=ln2−2+(π/2) 3λ=ln2+(π/(√3)) lnL=ln2−2+(π/2) 3λ−lnL=2+π((1/(√3))−(1/2)) pls check...$
	$L = \lim_{n \to \infty} {[\underset{r = 1}{\prod^{n}} \frac{(n^{2} + r^{2})}{n^{2}}]}^{\frac{1}{n}}$ $= \lim_{n \to \infty} {[{1 + {(\frac{1}{n})}^{2}} {1 + {(\frac{2}{n})}^{2}} {1 + {(\frac{3}{n})}^{2}} . . . {1 + {(\frac{n}{n})}^{2}}]}^{\frac{1}{n}}$ $h = \frac{1}{n}$ $l n L = \lim_{h \to 0} h [l n (1 + 1^{2} h^{2}) + l n (1 + 2^{2} h^{2}) + l n (1 + 3^{2} h^{2}) + . . . + l n (1 + n^{2} h^{2})]$ $l n L = \int_{0}^{1} l n (1 + x^{2}) d x$ $I_{1} = \int l n (1 + x^{2}) d x$ $= l n (1 + x^{2}) \times x - \int \frac{2 x \times x}{1 + x^{2}} d x$ $= x l n (1 + x^{2}) - 2 \int \frac{1 + x^{2} - 1}{1 + x^{2}} d x$ $= x l n (1 + x^{2}) - 2 \int (1 - \frac{1}{1 + x^{2}}) d x$ $= x l n (1 + x^{2}) - 2 [x - {t a n}^{- 1} x]$ $l n L =∣ x l n (1 + x^{2}) - 2 x + 2 {t a n}^{- 1} x ∣_{0}^{1}$ $l n L = [{l n 2 - 2 + 2 {t a n}^{- 1} 1}]$ $l n L = l n 2 - 2 + 2 \times \frac{π}{4}$ $l n L = l n 2 - 2 + \frac{π}{2}$ $3 λ = l n 2 + \frac{π}{\sqrt{3}} l n L = l n 2 - 2 + \frac{π}{2}$ $3 λ - l n L = 2 + π (\frac{1}{\sqrt{3}} - \frac{1}{2}) p l s c h e c k . . .$
	Commented by rahul 19 last updated on 08/Oct/18
	thanks sir!☺️
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