calculate ∫_0 ^∞ ((ln(x))/(x^2 +x+1))dx .


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Question Number 55457 by maxmathsup by imad last updated on 24/Feb/19

$c a l c u l a t e \int_{0}^{\infty} \frac{l n (x)}{x^{2} + x + 1} d x .$
Commented by maxmathsup by imad last updated on 05/Mar/19
$let I =∫_0 ^∞ ((ln(x))/(x^2 +x+1))dx ⇒I =∫_0 ^∞ ((lnx)/((x+(1/2))^2 +(3/4)))dx =_(x+(1/2)=((√3)/2)t) ∫_(1/(√3)) ^(+∞) ((ln(((√3)/2)t−(1/2)))/((3/4)(1+t^2 ))) ((√3)/2)dt =(4/3) ((√3)/2) ∫_(1/(√3)) ^(+∞) ((ln(((√3)/2)t−(1/2)))/(1+t^2 ))dt =(2/(√3)) ∫_(1/(√3)) ^(+∞) ((ln(((√3)/2)t−(1/2)))/(t^2 +1))dt ⇒ I =(2/(√3)) ∫_(1/(√3)) ^(+∞) ((ln(((√3)/2)x−(1/2)))/(x^2 +1))dx let f(t) =∫_(1/(√3)) ^(+∞) ((ln(tx−1))/(x^2 +1))dx ⇒f^′ (t) =∫_(1/(√3)) ^(+∞) (x/((tx−1)(x^2 +1)))dx let decompise F(x) =(x/((tx−1)(x^2 +1))) ⇒F(x)=(a/(tx−1)) +((bx +c)/(x^2 +1)) a =lim_(x→(1/t)) (tx−1)F(x) =(1/(t((1/t^2 ) +1))) =(1/(t+(1/t))) =(t/(t^2 +1)) lim_(x→+∞) xF(x)=(a/t) +b =0 ⇒b =−(a/t) =−(1/(t^2 +1)) ⇒ F(x) =(t/((t^2 +1)(tx−1))) +((−(1/(t^2 +1))x+c)/(x^2 +1)) F(0) =0 =−(t/(t^2 +1)) +c ⇒c =(t/(t^2 +1)) ⇒ F(x) =(t/((t^2 +1)(tx−1))) +(1/(t^2 +1)) ((−x+t)/(x^2 +1)) ⇒ ∫ F(x)dx =(1/(t^2 +1))ln∣tx−1∣−(1/(2(t^2 +1)))ln(x^2 +1) +(t/(t^2 +1)) arctan(x) +c ⇒ ∫_(1/(√3)) ^(+∞) F(x)dx =(1/(t^2 +1))[ln∣((tx−1)/(√(x^2 +1)))∣]_(1/(√3)) ^(+∞) +(t/(t^2 +1))[arctant]_(1/(√3)) ^(+∞) =(1/(t^2 +1)){ln∣t∣−ln∣(((t/(√3))−1)/(2/(√3)))∣} +(t/(t^2 +1)){(π/2) −(π/6)} =(1/(t^2 +1)){ln∣t∣−ln∣((t−(√3))/2)∣} +((πt)/(3(t^2 +1))) =f^′ (t) ⇒ f(t) =∫ ((ln∣t∣)/(t^2 +1))dt−∫ ((ln∣t−(√3))−ln(2))/(t^2 +1)) dt +(π/6)ln(t^2 +1) +C =∫ ((ln∣t∣)/(t^2 +1))dt −∫((ln∣t−(√3)∣)/(t^2 +1))dt+ln(2)arctant +C ....be continued....$
$l e t I = \int_{0}^{\infty} \frac{l n (x)}{x^{2} + x + 1} d x \Rightarrow I = \int_{0}^{\infty} \frac{l n x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} d x$ $=_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (\frac{\sqrt{3}}{2} t - \frac{1}{2})}{\frac{3}{4} (1 + t^{2})} \frac{\sqrt{3}}{2} d t = \frac{4}{3} \frac{\sqrt{3}}{2} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (\frac{\sqrt{3}}{2} t - \frac{1}{2})}{1 + t^{2}} d t$ $= \frac{2}{\sqrt{3}} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (\frac{\sqrt{3}}{2} t - \frac{1}{2})}{t^{2} + 1} d t \Rightarrow I = \frac{2}{\sqrt{3}} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (\frac{\sqrt{3}}{2} x - \frac{1}{2})}{x^{2} + 1} d x l e t$ $f (t) = \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (t x - 1)}{x^{2} + 1} d x \Rightarrow f^{^{'}} (t) = \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{x}{(t x - 1) (x^{2} + 1)} d x l e t d e c o m p i s e$ $F (x) = \frac{x}{(t x - 1) (x^{2} + 1)} \Rightarrow F (x) = \frac{a}{t x - 1} + \frac{b x + c}{x^{2} + 1}$ $a = {l i m}_{x \to \frac{1}{t}} (t x - 1) F (x) = \frac{1}{t (\frac{1}{t^{2}} + 1)} = \frac{1}{t + \frac{1}{t}} = \frac{t}{t^{2} + 1}$ ${l i m}_{x \to + \infty} x F (x) = \frac{a}{t} + b = 0 \Rightarrow b = - \frac{a}{t} = - \frac{1}{t^{2} + 1} \Rightarrow$ $F (x) = \frac{t}{(t^{2} + 1) (t x - 1)} + \frac{- \frac{1}{t^{2} + 1} x + c}{x^{2} + 1}$ $F (0) = 0 = - \frac{t}{t^{2} + 1} + c \Rightarrow c = \frac{t}{t^{2} + 1} \Rightarrow$ $F (x) = \frac{t}{(t^{2} + 1) (t x - 1)} + \frac{1}{t^{2} + 1} \frac{- x + t}{x^{2} + 1} \Rightarrow$ $\int F (x) d x = \frac{1}{t^{2} + 1} l n ∣ t x - 1 ∣ - \frac{1}{2 (t^{2} + 1)} l n (x^{2} + 1) + \frac{t}{t^{2} + 1} a r c t a n (x) + c \Rightarrow$ $\int_{\frac{1}{\sqrt{3}}}^{+ \infty} F (x) d x = \frac{1}{t^{2} + 1} {[l n ∣ \frac{t x - 1}{\sqrt{x^{2} + 1}} ∣]}_{\frac{1}{\sqrt{3}}}^{+ \infty} + \frac{t}{t^{2} + 1} {[a r c t a n t]}_{\frac{1}{\sqrt{3}}}^{+ \infty}$ $= \frac{1}{t^{2} + 1} {l n ∣ t ∣ - l n ∣ \frac{\frac{t}{\sqrt{3}} - 1}{\frac{2}{\sqrt{3}}} ∣} + \frac{t}{t^{2} + 1} {\frac{π}{2} - \frac{π}{6}}$ $= \frac{1}{t^{2} + 1} {l n ∣ t ∣ - l n ∣ \frac{t - \sqrt{3}}{2} ∣} + \frac{π t}{3 (t^{2} + 1)} = f^{^{'}} (t) \Rightarrow$ $f (t) = \int \frac{l n ∣ t ∣}{t^{2} + 1} d t - \int \frac{l n ∣ t - \sqrt{3}) - l n (2)}{t^{2} + 1} d t + \frac{π}{6} l n (t^{2} + 1) + C$ $= \int \frac{l n ∣ t ∣}{t^{2} + 1} d t - \int \frac{l n ∣ t - \sqrt{3} ∣}{t^{2} + 1} d t + l n (2) a r c t a n t + C . . . . b e c o n t i n u e d . . . .$
Commented by Abdo msup. last updated on 05/Mar/19

$l e t t r y a n o t h e r w a y w e h a v e I = \int_{0}^{\infty} \frac{(1 - x) l n x}{1 - x^{3}} d x$ $= \int_{0}^{1} \frac{(1 - x) l n (x)}{1 - x^{3}} d x + \int_{1}^{+ \infty} \frac{(1 - x) l n (x)}{1 - x^{3}} d x b u t$ $\int_{1}^{\infty} \frac{(1 - x) l n (x)}{1 - x^{3}} d x =_{x = \frac{1}{t}} - \int_{0}^{1} \frac{(1 - \frac{1}{t}) (- l n (t))}{1 - \frac{1}{t^{3}}} (\frac{- d t}{t^{2}})$ $= - \int_{0}^{1} \frac{(t - 1) l n (t)}{t^{3} - 1} d t = \int_{0}^{1} \frac{(t - 1) l n (t)}{1 - t^{3}} d t$ $= - \int_{0}^{1} \frac{(1 - t) l n (t)}{1 - t^{3}} d t \Rightarrow I = 0$
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