Evaluate ∫_0 ^1 ((Log(x))/(x^2 +2x+3)) dx


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Question Number 48289 by Abdulhafeez Abu qatada last updated on 21/Nov/18

$E v a l u a t e \underset{0}{\int^{1}} \frac{L o g (x)}{x^{2} + 2 x + 3} d x$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Nov/18

Commented by maxmathsup by imad last updated on 22/Nov/18
$let A =∫_0 ^1 ((ln(x))/(x^2 +2x+3))dx ⇒A =∫_0 ^1 ((ln(x))/((x+1)^2 +2)) A =_(x+1 =(√2)t) ∫_(1/(√2)) ^(√2) ((ln(t(√2)−1))/(2(1+t^2 ))) (√2)dt =(1/(√2)) ∫_(1/(√2)) ^(√2) ((ln(t(√2)−1))/(1+t^2 ))dt let consider the parametric function ϕ(x)=∫_(1/(√2)) ^(√2) ((ln(xt−1))/(1+t^2 ))dt (xt>1) ϕ^′ (x) = ∫_(1/(√2)) ^(√2) (t/((xt−1)(1+t^2 ))) dt let decompose F(t)=(t/((xt−1)(1+t^2 ))) F(t)=(a/(xt−1)) +((bt+c)/(t^2 +1)) a =lim_(t→(1/x)) (xt−1)F(t) =(1/(x(1+(1/x^2 )))) =(1/(x+(1/x))) =(x/(x^2 +1)) lim_(t→+∞) t F(t)=(a/x) +b ⇒b=−(1/(x^2 +1)) ⇒F(t)=(x/((x^2 +1)(xt−1))) +((−(t/(x^2 +1))+c)/(t^2 +1)) F(0) =0 =−(x/((x^2 +1))) +c ⇒c=(x/(x^2 +1)) ⇒ F(t)= (x/((x^2 +1)(xt−1))) −(1/(x^2 +1)) ((t −x)/(t^2 +1)) ⇒ ϕ^′ (x) =∫_(1/(√2)) ^(√2) ( (x/((x^2 +1)(xt−1))) −(1/(x^2 +1)) ((t−x)/(t^2 +1)))dt =(x/(x^2 +1)) ∫_(1/(√2)) ^(√2) (dt/(xt−1)) −(1/(2(x^2 +1))) ∫_(1/(√2)) ^(√2) ((2t)/(t^2 +1)) +(x/(x^2 +1)) ∫_(1/(√2)) ^(√2) (dt/(t^2 +1)) =(1/(x^2 +1))ln∣xt−1∣]_(1/(√2)) ^(√2) −(1/(2(x^2 +1)))[ln(t^2 +1)]_(1/(√2)) ^(√2) +(x/(x^2 +1)) [arctan(t)]_(1/(√2)) ^(√2) =(1/(x^2 +1)){ln∣x(√2)−1∣−ln∣(x/(√2))−1∣}−((ln(2))/(2(x^2 +1))) +(x/(x^2 +1)) {arctan((√2))−arctan((1/(√2)))} ⇒ ϕ(x) = ∫ ((ln(x(√2)−1))/(x^2 +1)) dx−∫ ((ln((x/(√2))−1))/(x^2 +1))dx −((ln(2))/2) ∫ (dx/(1+x^2 )) +(arctan((√2))−arctan((1/(√2))) ∫ ((xdx)/(x^2 +1))+c ⇒ ϕ(x) =∫ ((ln(x(√2)−1))/(x^2 +1))dx−∫ ((ln(x−(√2)))/(x^2 +1)) dx +ln((√2))arctanx−((ln(2))/2) arctanx +(1/2)(arctan((√2))−arctan((1/(√2))))ln(1+x^2 )+c =∫ ((ln(x(√2)−1))/(x^2 +1))dx−∫ ((ln(x−(√2)))/(x^2 +1))dx +(1/2)(arctan((√2))−arctan((1/(√2))))ln(1+x^2 )+c ....be continued...$
$l e t A = \int_{0}^{1} \frac{l n (x)}{x^{2} + 2 x + 3} d x \Rightarrow A = \int_{0}^{1} \frac{l n (x)}{{(x + 1)}^{2} + 2}$ $A =_{x + 1 = \sqrt{2} t} \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \frac{l n (t \sqrt{2} - 1)}{2 (1 + t^{2})} \sqrt{2} d t = \frac{1}{\sqrt{2}} \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \frac{l n (t \sqrt{2} - 1)}{1 + t^{2}} d t l e t c o n s i d e r t h e$ $p a r a m e t r i c f u n c t i o n φ (x) = \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \frac{l n (x t - 1)}{1 + t^{2}} d t (x t > 1)$ $φ^{^{'}} (x) = \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \frac{t}{(x t - 1) (1 + t^{2})} d t l e t d e c o m p o s e F (t) = \frac{t}{(x t - 1) (1 + t^{2})}$ $F (t) = \frac{a}{x t - 1} + \frac{b t + c}{t^{2} + 1}$ $a = {l i m}_{t \to \frac{1}{x}} (x t - 1) F (t) = \frac{1}{x (1 + \frac{1}{x^{2}})} = \frac{1}{x + \frac{1}{x}} = \frac{x}{x^{2} + 1}$ ${l i m}_{t \to + \infty} t F (t) = \frac{a}{x} + b \Rightarrow b = - \frac{1}{x^{2} + 1} \Rightarrow F (t) = \frac{x}{(x^{2} + 1) (x t - 1)} + \frac{- \frac{t}{x^{2} + 1} + c}{t^{2} + 1}$ $F (0) = 0 = - \frac{x}{(x^{2} + 1)} + c \Rightarrow c = \frac{x}{x^{2} + 1} \Rightarrow$ $F (t) = \frac{x}{(x^{2} + 1) (x t - 1)} - \frac{1}{x^{2} + 1} \frac{t - x}{t^{2} + 1} \Rightarrow$ $φ^{^{'}} (x) = \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} (\frac{x}{(x^{2} + 1) (x t - 1)} - \frac{1}{x^{2} + 1} \frac{t - x}{t^{2} + 1}) d t$ $= \frac{x}{x^{2} + 1} \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \frac{d t}{x t - 1} - \frac{1}{2 (x^{2} + 1)} \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \frac{2 t}{t^{2} + 1} + \frac{x}{x^{2} + 1} \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \frac{d t}{t^{2} + 1}$ ${= \frac{1}{x^{2} + 1} l n ∣ x t - 1 ∣]}_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} - \frac{1}{2 (x^{2} + 1)} {[l n (t^{2} + 1)]}_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} + \frac{x}{x^{2} + 1} {[a r c t a n (t)]}_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}$ $= \frac{1}{x^{2} + 1} {l n ∣ x \sqrt{2} - 1 ∣ - l n ∣ \frac{x}{\sqrt{2}} - 1 ∣} - \frac{l n (2)}{2 (x^{2} + 1)}$ $+ \frac{x}{x^{2} + 1} {a r c t a n (\sqrt{2}) - a r c t a n (\frac{1}{\sqrt{2}})} \Rightarrow$ $φ (x) = \int \frac{l n (x \sqrt{2} - 1)}{x^{2} + 1} d x - \int \frac{l n (\frac{x}{\sqrt{2}} - 1)}{x^{2} + 1} d x - \frac{l n (2)}{2} \int \frac{d x}{1 + x^{2}}$ $+ (a r c t a n (\sqrt{2}) - a r c t a n (\frac{1}{\sqrt{2}}) \int \frac{x d x}{x^{2} + 1} + c \Rightarrow$ $φ (x) = \int \frac{l n (x \sqrt{2} - 1)}{x^{2} + 1} d x - \int \frac{l n (x - \sqrt{2})}{x^{2} + 1} d x + l n (\sqrt{2}) a r c t a n x - \frac{l n (2)}{2} a r c t a n x$ $+ \frac{1}{2} (a r c t a n (\sqrt{2}) - a r c t a n (\frac{1}{\sqrt{2}})) l n (1 + x^{2}) + c$ $= \int \frac{l n (x \sqrt{2} - 1)}{x^{2} + 1} d x - \int \frac{l n (x - \sqrt{2})}{x^{2} + 1} d x + \frac{1}{2} (a r c t a n (\sqrt{2}) - a r c t a n (\frac{1}{\sqrt{2}})) l n (1 + x^{2}) + c$ $. . . . b e c o n t i n u e d . . .$
Commented by maxmathsup by imad last updated on 22/Nov/18

$w e h a v e 0 < x^{2} < 1, 0 < 2 x < 2 \Rightarrow 3 < x^{2} + 2 x + 3 < 5 \Rightarrow \frac{1}{5} < \frac{1}{x^{2} + 2 x + 3} < \frac{1}{3} \Rightarrow$ $\Rightarrow - \frac{1}{5} l n (x) < - \frac{l n x}{x^{2} + 2 x + 3} < - \frac{1}{3} l n (x) (s e e t h a t l n (x) < 0 o n] 0, 1 [) \Rightarrow$ $\frac{1}{3} l n (x) < \frac{l n (x)}{x^{2} + 2 x + 3} < \frac{1}{5} l n (x) \Rightarrow \frac{1}{3} \int_{0}^{1} l n (x) d x < \int_{0}^{1} \frac{l n (x)}{x^{2} + 2 x + 3} d x < \frac{1}{5} \int_{0}^{1} l n (x) \Rightarrow$ $\frac{1}{3} {[x l n (x) - x]}_{0}^{1} < \int_{0}^{1} \frac{l n (x)}{x^{2} + 2 x + 3} < \frac{1}{5} {[x l n (x) - x]}_{0}^{1} \Rightarrow$ $- \frac{1}{3} < \int_{0}^{1} \frac{l n (x)}{x^{2} + 2 x + 3} d x < - \frac{1}{5} w e c a n t a k e \int_{0}^{1} \frac{l n (x)}{x^{2} + 2 x + 3} \sim - \frac{8}{30}$ $w i t h p r e c i s i o n σ = \frac{\frac{1}{3} - \frac{1}{5}}{2} = \frac{2}{30} .$
Commented by maxmathsup by imad last updated on 22/Nov/18

$s o \int_{0}^{1} \frac{l n (x)}{x^{2} + 2 x + 3} d x \sim - 0, 25 .$
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