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


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Question Number 73178 by mathmax by abdo last updated on 07/Nov/19

$c a l c u l a t e \int_{0}^{\infty} \frac{l n (2 + x^{2})}{x^{2} - x + 1} d x$
	Answered by mind is power last updated on 07/Nov/19
	$=∫_∞ ^0 ((ln(2+(1/x^2 )))/((1/x^2 )−(1/x)+1)).−(dx/x^2 )=∫_0 ^(+∞) ((ln(2x^2 +1)−ln(x^2 ))/(1−x+x^2 ))dx =∫_0 ^(+∞) ((ln(2x^2 +1))/(1−x+x^2 ))dx−2∫_0 ^(+∞) ((ln(x))/(1−x+x^2 )) ∫_0 ^(+∞) ((ln(x))/(1−x+x^2 ))dx=−∫_0 ^(+∞) ((ln(x))/(1−x+x^2 ))dx⇒=0 =∫_0 ^(+∞) ((ln(2x^2 +1))/(1−x+x^2 ))dx let f(t)=∫_0 ^(+∞) ((ln(tx^2 +1))/(1−x+x^2 ))dx t≥0 f(0)=0 f′(t)=∫_0 ^∞ ((x^2 dx)/((tx^2 +1)(1−x+x^2 ))) f′(t)=∫_0 ^(+∞) ((ax+b)/(tx^2 +1))+((cx+d)/(1−x+x^2 )) b+d=0 (a/t)+c=0⇒a+tc=0 a−b+c=0 dt−a+b=1 b=−d a=−tc −tc+d+c=0 dt+tc−d=1 d=c(t−1) c(t−1)^2 +tc=1 c=(1/(t^2 −t+1)),d=((t−1)/(t^2 −t+1)),a=((−t)/(t^2 −t+1)),b=((1−t)/(t^2 −t+1)) f′(t)=(1/(t^2 −t+1))∫_0 ^(+∞) ((−tx+(1−t))/(tx^2 +1))dx+(1/(t^2 −t+1))∫_0 ^(+∞) ((x+t−1)/(x^2 −x+1)) =(1/(t^2 −t+1))∫_0 ^(+∞) ((−tx)/(tx^2 +1))+((1−t)/(t^2 −t+1))∫_0 ^(+∞) (dx/(tx^2 +1))+(1/(t^2 −t+1))∫_0 ^(+∞) ((x−(1/2))/(x^2 −x+1))dx +((t−(1/2))/(t^2 −t+1))∫_0 ^(+∞) (dx/(x^2 −x+1)) =lim_(y→∞) ((1/2)/(t^2 −t+1))ln(((x^2 −x+1)/(tx^2 +1)))_0 ^y =−((ln(t))/(2(t^2 −t+1))) ∫_0 ^(+∞) (dx/(tx^2 +1))=(1/(√t)).(π/2) ∫_0 ^(+∞) (dx/(x^2 −x+1))=∫_0 ^(+∞) (dx/((x−(1/2))^2 +(3/4)))=(2/(√3)).{(π/2)+(π/6)}=((4π)/(3(√3))) ⇒f′(t)=((ln(t))/(2(t^2 −t+1)))+((1−t)/(t^2 −t+1)).(π/2)(√t)+((t−(1/2))/(t^2 −t+1)).((4π)/(3(√3))) ⇒our integral is f(2),f(0)=0 ∫_0 ^(+∞) ((ln(2x^2 +1))/(x^2 −x+1))dx=f(2)=∫_0 ^2 f′(t)dt =∫_0 ^2 ((ln(t)dt)/(2(t^2 −t+1)))+(π/2)∫_0 ^2 (((1−t)(√t))/(t^2 −t+1))dt+((4π)/(3(√3)))∫_0 ^2 ((2t−1)/(2(t^2 −t+1)))dt only ∫_0 ^2 ((ln(t))/(t^2 −t+1))dt is tricky t^2 −t+1=(t−((1−i(√3))/2))(t−((1+i(√3))/2)) (1/(t^2 −t+1))=((−i(√3))/(t−((1−i(√3))/2)))+((i(√3))/(t−((1+i(√3))/2))) ∫_0 ^2 ((ln(t))/(t^2 −t+1))dt=−i(√3)∫_0 ^2 ((ln(t))/(t−((1−i(√3))/2)))dt+i(√3)∫_0 ^2 (dt/(t−((1+i(√3))/2)))=c j=((1−i(√3))/2),f=((1+i(√3))/2) c=i(√3)∫_0 ^2 ((ln(t))/(1−ft))dt−i(√3)∫_0 ^2 ((ln(t))/(1−jt))dt u=ft in firt ,jt=s in 2nd c=i(√3)j∫_0 ^(2f) ((ln(u)+ln(j))/(1−u))du−i(√3)f∫_0 ^(2j) ((ln(s)+ln(f))/(1−s))ds ∫_1 ^(1−z) ((ln(t))/(1−t))dt=Li_2 (z) ⇒∫_0 ^(2f) ((ln(u)+ln(j))/(1−u))du=Li_2 (1−2f)−Li_2 (1)−log(j)log(1−2f) ∫_0 ^(2j) ((ln(s)+ln(f))/(1−s))=Li_2 (1−2j)−Li_2 (1)−log(f)log(1−2j) ⇒∫_0 ^2 ((ln(t))/(t^2 −t+1))dt=i(√3)(((1−i(√3))/2)){Li_2 (−i(√3))−Li_2 (1)+((iπ)/3)Log(−i(√3)} −i(√3)(((1+i(√3))/2)){Li_2 (i(√3))−Li_2 (1)−((iπ)/3)log(i(√3))}$
	$= \int_{\infty}^{0} \frac{\ln (2 + \frac{1}{x^{2}})}{\frac{1}{x^{2}} - \frac{1}{x} + 1} . - \frac{dx}{x^{2}} = \int_{0}^{+ \infty} \frac{\ln ({2 x}^{2} + 1) - \ln (x^{2})}{1 - x + x^{2}} dx$ $= \int_{0}^{+ \infty} \frac{\ln ({2 x}^{2} + 1)}{1 - x + x^{2}} dx - 2 \int_{0}^{+ \infty} \frac{\ln (x)}{1 - x + x^{2}}$ $\int_{0}^{+ \infty} \frac{\ln (x)}{1 - x + x^{2}} dx = - \int_{0}^{+ \infty} \frac{\ln (x)}{1 - x + x^{2}} dx \Rightarrow= 0$ $= \int_{0}^{+ \infty} \frac{\ln ({2 x}^{2} + 1)}{1 - x + x^{2}} dx let f (t) = \int_{0}^{+ \infty} \frac{\ln ({tx}^{2} + 1)}{1 - x + x^{2}} dx t ⩾ 0$ $f (0) = 0$ $f^{'} (t) = \int_{0}^{\infty} \frac{x^{2} dx}{({tx}^{2} + 1) (1 - x + x^{2})}$ $f^{'} (t) = \int_{0}^{+ \infty} \frac{ax + b}{{tx}^{2} + 1} + \frac{cx + d}{1 - x + x^{2}}$ $b + d = 0$ $\frac{a}{t} + c = 0 \Rightarrow a + tc = 0$ $a - b + c = 0$ $dt - a + b = 1$ $b = - d$ $a = - tc$ $- tc + d + c = 0$ $dt + tc - d = 1$ $d = c (t - 1)$ $c {(t - 1)}^{2} + tc = 1$ $c = \frac{1}{t^{2} - t + 1}, d = \frac{t - 1}{t^{2} - t + 1}, a = \frac{- t}{t^{2} - t + 1}, b = \frac{1 - t}{t^{2} - t + 1}$ $f^{'} (t) = \frac{1}{t^{2} - t + 1} \int_{0}^{+ \infty} \frac{- tx + (1 - t)}{{tx}^{2} + 1} dx + \frac{1}{t^{2} - t + 1} \int_{0}^{+ \infty} \frac{x + t - 1}{x^{2} - x + 1}$ $= \frac{1}{t^{2} - t + 1} \int_{0}^{+ \infty} \frac{- tx}{{tx}^{2} + 1} + \frac{1 - t}{t^{2} - t + 1} \int_{0}^{+ \infty} \frac{dx}{{tx}^{2} + 1} + \frac{1}{t^{2} - t + 1} \int_{0}^{+ \infty} \frac{x - \frac{1}{2}}{x^{2} - x + 1} dx$ $+ \frac{t - \frac{1}{2}}{t^{2} - t + 1} \int_{0}^{+ \infty} \frac{dx}{x^{2} - x + 1}$ $= \lim_{y \to \infty} \frac{\frac{1}{2}}{t^{2} - t + 1} \ln {(\frac{x^{2} - x + 1}{{tx}^{2} + 1})}_{0}^{y} = - \frac{\ln (t)}{2 (t^{2} - t + 1)}$ $\int_{0}^{+ \infty} \frac{dx}{{tx}^{2} + 1} = \frac{1}{\sqrt{t}} . \frac{π}{2}$ $\int_{0}^{+ \infty} \frac{dx}{x^{2} - x + 1} = \int_{0}^{+ \infty} \frac{dx}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} = \frac{2}{\sqrt{3}} . {\frac{π}{2} + \frac{π}{6}} = \frac{4 π}{3 \sqrt{3}}$ $\Rightarrow f^{'} (t) = \frac{\ln (t)}{2 (t^{2} - t + 1)} + \frac{1 - t}{t^{2} - t + 1} . \frac{π}{2} \sqrt{t} + \frac{t - \frac{1}{2}}{t^{2} - t + 1} . \frac{4 π}{3 \sqrt{3}}$ $\Rightarrow our integral is f (2), f (0) = 0$ $\int_{0}^{+ \infty} \frac{\ln ({2 x}^{2} + 1)}{x^{2} - x + 1} dx = f (2) = \int_{0}^{2} f^{'} (t) dt$ $= \int_{0}^{2} \frac{\ln (t) dt}{2 (t^{2} - t + 1)} + \frac{π}{2} \int_{0}^{2} \frac{(1 - t) \sqrt{t}}{t^{2} - t + 1} dt + \frac{4 π}{3 \sqrt{3}} \int_{0}^{2} \frac{2 t - 1}{2 (t^{2} - t + 1)} dt$ $only \int_{0}^{2} \frac{\ln (t)}{t^{2} - t + 1} dt is tricky$ $t^{2} - t + 1 = (t - \frac{1 - i \sqrt{3}}{2}) (t - \frac{1 + i \sqrt{3}}{2})$ $\frac{1}{t^{2} - t + 1} = \frac{- i \sqrt{3}}{t - \frac{1 - i \sqrt{3}}{2}} + \frac{i \sqrt{3}}{t - \frac{1 + i \sqrt{3}}{2}}$ $\int_{0}^{2} \frac{\ln (t)}{t^{2} - t + 1} dt = - i \sqrt{3} \int_{0}^{2} \frac{\ln (t)}{t - \frac{1 - i \sqrt{3}}{2}} dt + i \sqrt{3} \int_{0}^{2} \frac{dt}{t - \frac{1 + i \sqrt{3}}{2}} = c$ $j = \frac{1 - i \sqrt{3}}{2}, f = \frac{1 + i \sqrt{3}}{2}$ $c = i \sqrt{3} \int_{0}^{2} \frac{\ln (t)}{1 - ft} dt - i \sqrt{3} \int_{0}^{2} \frac{\ln (t)}{1 - jt} dt$ $u = ft in firt, jt = s in 2 nd$ $c = i \sqrt{3} j \int_{0}^{2 f} \frac{\ln (u) + \ln (j)}{1 - u} du - i \sqrt{3} f \int_{0}^{2 j} \frac{\ln (s) + \ln (f)}{1 - s} ds$ $\int_{1}^{1 - z} \frac{\ln (t)}{1 - t} dt = {Li}_{2} (z)$ $\Rightarrow \int_{0}^{2 f} \frac{\ln (u) + \ln (j)}{1 - u} du = {Li}_{2} (1 - 2 f) - {Li}_{2} (1) - \log (j) \log (1 - 2 f)$ $\int_{0}^{2 j} \frac{\ln (s) + \ln (f)}{1 - s} = {Li}_{2} (1 - 2 j) - {Li}_{2} (1) - \log (f) \log (1 - 2 j)$ $\Rightarrow \int_{0}^{2} \frac{\ln (t)}{t^{2} - t + 1} dt = i \sqrt{3} (\frac{1 - i \sqrt{3}}{2}) {{Li}_{2} (- i \sqrt{3}) - {Li}_{2} (1) + \frac{i π}{3} Log (- i \sqrt{3}}$ $- i \sqrt{3} (\frac{1 + i \sqrt{3}}{2}) {{Li}_{2} (i \sqrt{3}) - {Li}_{2} (1) - \frac{i π}{3} \log (i \sqrt{3})}$
	Commented by mathmax by abdo last updated on 07/Nov/19

	$t h a n k x s i r .$
	Commented by mind is power last updated on 07/Nov/19

	$y^{'} re welcom sir$
	Commented by aliesam last updated on 15/Dec/19

	$I {d i d n}^{'} t g e t t h e f i r s t s t e p w h e r e y o u f a c t o r - \frac{d x}{x^{2}}$ $i m e a n t h e l n p a r t$
	Commented by aliesam last updated on 15/Dec/19

	$I m e a n$ $\int_{0}^{\infty} \frac{l n (2 + x^{2})}{x^{2} - x + 1} = \int_{\infty}^{0} \frac{l n (2 + \frac{1}{x^{2}})}{\frac{1}{x^{2}} - \frac{1}{x} + 1} . \frac{- d x}{x^{2}} ? ?$
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