∫ln(x+1)/(x^2 −x+1) limit ={ 0>2}


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Question Number 62252 by hovea cw last updated on 18/Jun/19
$∫ln(x+1)/(x^2 −x+1) limit ={ 0>2}$
$\int \ln (x + 1) / (x^{2} - x + 1)$ $limit = {0 > 2}$
Commented bymaxmathsup by imad last updated on 18/Jun/19
$let I =∫_0 ^2 ((ln(x+1))/(x^2 −x +1))dx and f(t)=∫_0 ^2 ((ln(1+tx))/(x^2 −x +1)) dx with t≥0 we have I =f(1) and f^′ (t)=∫_0 ^2 (x/((1+tx)(x^2 −x+1)))dx =(1/t)∫_0 ^2 ((1+tx−1)/((1+tx)(x^2 −x+1)))dx =(1/t)∫_0 ^2 (dx/((x^2 −x+1))) −(1/t) ∫_0 ^2 (dx/((tx+1)(x^2 −x+1))) ∫_0 ^2 (dx/(x^2 −x+1)) =∫_0 ^2 (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)u) (4/3) ∫_(−(1/(√3))) ^(3/(√3)) (1/(u^2 +1)) ((√3)/2)du =(2/(√3)) [arctn(u)]_(−(1/(√3))) ^(√3) =(2/(√3)){ arctan((√3))+arctan((1/(√3)))}=(2/(√3))(π/2) =(π/(√3)) let decompose F(x)=(1/((tx+1)(x^2 −x +1)→(Δ<0))) F(x) =(a/(tx+1)) +((bx+c)/(x^2 −x+1)) a =lim_(x→−(1/t)) (tx+1)F(x) =(t^2 /(t^2 +t+1)) ⇒F(x) =(t^2 /((tx+1)(t^2 +t+1))) +((bx+c)/(x^2 −x +1)) lim_(t→+∞) xF(x) =0 =(a/t) +b ⇒a+tb =0 ⇒tb =−a ⇒b =−(a/t) =−(t/(t^2 +t+1)) ⇒ F(x) =(t^2 /((t^2 +t+1)(tx+1))) +((−(t/(t^2 +t+1))x +c)/(x^2 −x+1)) F(0)=1 =a+c ⇒c =1−a =1−(t^2 /(t^2 +t+1)) =((t+1)/(t^2 +t+1)) ⇒ F(x) =(t^2 /((t^2 +t+1)(tx+1))) −(1/(t^2 +t+1)) ((tx−t−1)/(x^2 −x+1)) ⇒ ∫_0 ^2 F(x)dx =(t^2 /(t^2 +t+1)) ∫_0 ^2 (dx/(tx+1)) −(1/(t^2 +t+1)) ∫_0 ^2 ((tx−t−1)/(x^2 −x +1))dx ∫_0 ^2 (dx/(tx +1)) =(1/t)[ln∣tx+1∣]_0 ^2 =(1/t)ln∣2t+1∣ ∫_0 ^2 ((tx−t−1)/(x^2 −x+1))dx =(t/2)∫_0 ^2 ((2x−1+1)/(x^2 −x+1))dx −(t+1)∫_0 ^2 (dx/(x^2 −x+1)) =(t/2)[ln∣x^2 −x+1∣]_0 ^2 −(t+1)(π/(√3)) =(t/2)ln(3)−(π/(√3))(t+1) ⇒ ∫_0 ^2 F(x)dx =(t/(t^2 +t +1))ln∣2t+1∣ −(1/(t^2 +t+1)){(((ln(3))/2)−(π/(√3)))t−(π/(√3))} ⇒ f^′ (t) =(π/(t(√3))) −(1/t)∫_0 ^2 F(x)dx =(π/(t(√3))) −((ln∣2t+1∣)/(t^(2 ) +t +1)) −((((ln(3))/2)−(π/(√3)))/(t^2 +t +1)) +(π/(t(√3)(t^2 +t +1))) ⇒ f(t) =(π/(√3))ln(t)−∫ ((ln(2t+1))/(t^2 +t +1))dt −(((ln3)/2)−(π/(√3)))∫ (dt/(t^2 +t +1)) +(π/(√3)) ∫ (dt/(t(t^2 +t +1))) +c ...be continued...$
$l e t I = \int_{0}^{2} \frac{l n (x + 1)}{x^{2} - x + 1} d x a n d f (t) = \int_{0}^{2} \frac{l n (1 + t x)}{x^{2} - x + 1} d x w i t h t ⩾ 0$ $w e h a v e I = f (1) a n d f^{^{'}} (t) = \int_{0}^{2} \frac{x}{(1 + t x) (x^{2} - x + 1)} d x$ $= \frac{1}{t} \int_{0}^{2} \frac{1 + t x - 1}{(1 + t x) (x^{2} - x + 1)} d x = \frac{1}{t} \int_{0}^{2} \frac{d x}{(x^{2} - x + 1)} - \frac{1}{t} \int_{0}^{2} \frac{d x}{(t x + 1) (x^{2} - x + 1)}$ $\int_{0}^{2} \frac{d x}{x^{2} - x + 1} = \int_{0}^{2} \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} =_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{- \frac{1}{\sqrt{3}}}^{\frac{3}{\sqrt{3}}} \frac{1}{u^{2} + 1} \frac{\sqrt{3}}{2} d u$ $= \frac{2}{\sqrt{3}} {[a r c t n (u)]}_{- \frac{1}{\sqrt{3}}}^{\sqrt{3}} = \frac{2}{\sqrt{3}} {a r c t a n (\sqrt{3}) + a r c t a n (\frac{1}{\sqrt{3}})} = \frac{2}{\sqrt{3}} \frac{π}{2} = \frac{π}{\sqrt{3}}$ $l e t d e c o m p o s e F (x) = \frac{1}{(t x + 1) (x^{2} - x + 1) \to (Δ < 0)}$ $F (x) = \frac{a}{t x + 1} + \frac{b x + c}{x^{2} - x + 1}$ $a = {l i m}_{x \to - \frac{1}{t}} (t x + 1) F (x) = \frac{t^{2}}{t^{2} + t + 1} \Rightarrow F (x) = \frac{t^{2}}{(t x + 1) (t^{2} + t + 1)} + \frac{b x + c}{x^{2} - x + 1}$ ${l i m}_{t \to + \infty} x F (x) = 0 = \frac{a}{t} + b \Rightarrow a + t b = 0 \Rightarrow t b = - a \Rightarrow b = - \frac{a}{t} = - \frac{t}{t^{2} + t + 1} \Rightarrow$ $F (x) = \frac{t^{2}}{(t^{2} + t + 1) (t x + 1)} + \frac{- \frac{t}{t^{2} + t + 1} x + c}{x^{2} - x + 1}$ $F (0) = 1 = a + c \Rightarrow c = 1 - a = 1 - \frac{t^{2}}{t^{2} + t + 1} = \frac{t + 1}{t^{2} + t + 1} \Rightarrow$ $F (x) = \frac{t^{2}}{(t^{2} + t + 1) (t x + 1)} - \frac{1}{t^{2} + t + 1} \frac{t x - t - 1}{x^{2} - x + 1} \Rightarrow$ $\int_{0}^{2} F (x) d x = \frac{t^{2}}{t^{2} + t + 1} \int_{0}^{2} \frac{d x}{t x + 1} - \frac{1}{t^{2} + t + 1} \int_{0}^{2} \frac{t x - t - 1}{x^{2} - x + 1} d x$ $\int_{0}^{2} \frac{d x}{t x + 1} = \frac{1}{t} {[l n ∣ t x + 1 ∣]}_{0}^{2} = \frac{1}{t} l n ∣ 2 t + 1 ∣$ $\int_{0}^{2} \frac{t x - t - 1}{x^{2} - x + 1} d x = \frac{t}{2} \int_{0}^{2} \frac{2 x - 1 + 1}{x^{2} - x + 1} d x - (t + 1) \int_{0}^{2} \frac{d x}{x^{2} - x + 1}$ $= \frac{t}{2} {[l n ∣ x^{2} - x + 1 ∣]}_{0}^{2} - (t + 1) \frac{π}{\sqrt{3}} = \frac{t}{2} l n (3) - \frac{π}{\sqrt{3}} (t + 1) \Rightarrow$ $\int_{0}^{2} F (x) d x = \frac{t}{t^{2} + t + 1} l n ∣ 2 t + 1 ∣ - \frac{1}{t^{2} + t + 1} {(\frac{l n (3)}{2} - \frac{π}{\sqrt{3}}) t - \frac{π}{\sqrt{3}}} \Rightarrow$ $f^{^{'}} (t) = \frac{π}{t \sqrt{3}} - \frac{1}{t} \int_{0}^{2} F (x) d x$ $= \frac{π}{t \sqrt{3}} - \frac{l n ∣ 2 t + 1 ∣}{t^{2} + t + 1} - \frac{\frac{l n (3)}{2} - \frac{π}{\sqrt{3}}}{t^{2} + t + 1} + \frac{π}{t \sqrt{3} (t^{2} + t + 1)} \Rightarrow$ $f (t) = \frac{π}{\sqrt{3}} l n (t) - \int \frac{l n (2 t + 1)}{t^{2} + t + 1} d t - (\frac{l n 3}{2} - \frac{π}{\sqrt{3}}) \int \frac{d t}{t^{2} + t + 1} + \frac{π}{\sqrt{3}} \int \frac{d t}{t (t^{2} + t + 1)} + c$ $. . . b e c o n t i n u e d . . .$
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