let f(x) = ∫_0 ^1 ln(1+t +xt^2 )dt 1) calculate f^′ (x) then find a simple form of f(x) 2) calculate ∫_0 ^1 ln(1+t +t^2 )dt 3) calculate ∫


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Question Number 41679 by math khazana by abdo last updated on 11/Aug/18

$l e t f (x) = \int_{0}^{1} l n (1 + t + {x t}^{2}) d t$ $1) c a l c u l a t e f^{^{'}} (x) t h e n f i n d a s i m p l e f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{1} l n (1 + t + t^{2}) d t$ $3) c a l c u l a t e \int_{0}^{1} l n (1 - t^{3}) d t .$
Commented by maxmathsup by imad last updated on 11/Aug/18
$1) we have f^′ (x) = ∫_0 ^1 (t^2 /(xt^2 +t+1))dt =(1/x) ∫_0 ^1 ((xt^2 +t+1−t−1)/(xt^2 +t+1))dt =(1/x) −(1/x) ∫_0 ^1 ((t+1)/(xt^2 +t +1))dt let decompose F(t) =((t+1)/(xt^2 +t +1)) Δ =1−4x case 1 1−4x>0 ⇔ x<(1/4) ⇒t_1 =((−1+(√(1−4x)))/(2x)) and t_2 =((−1−(√(1−4x)))/(2x)) F(t)=((t+1)/(x(t−t_1 )(t−t_2 ))) = (a/(t−t_1 )) +(b/(t−t_2 )) a=((t_1 +1)/(x ((2(√(1−4x)))/(2x)))) = ((t_1 +1)/(√(1−4x))) b = ((t_2 +1)/(x(−((2(√(1−4x)))/(2x))))) =−((t_2 +1)/(√(1−4x))) ⇒ ∫_0 ^1 F(t)dt = ∫_0 ^1 {(a/(t−t_1 )) +(b/(t−t_2 ))}dt =[ aln∣t−t_1 ∣ +b ln∣t−t_2 ∣]_0 ^1 = aln∣1−t_1 ∣+bln∣1−t_2 ∣−aln∣t_1 ∣−bln∣t_2 ∣ =aln∣((1−t_1 )/t_1 )∣ +bln∣((1−t_2 )/t_2 )∣ ⇒ f^′ (x)=(1/x) −(1/x){aln∣((1−t_1 )/t_1 )∣+bln∣((1−t_2 )/t_2 )∣} ⇒ f(x)=(1−λ_0 )ln∣x∣ +c with λ_0 =aln∣((1−t_1 )/t_1 )∣+bln∣((1−t_2 )/t_2 )∣(known) c=f(1) ⇒ f(x)=(1−λ_0 )ln∣x∣ +f(1)$
$1) w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{t^{2}}{{x t}^{2} + t + 1} d t = \frac{1}{x} \int_{0}^{1} \frac{{x t}^{2} + t + 1 - t - 1}{{x t}^{2} + t + 1} d t$ $= \frac{1}{x} - \frac{1}{x} \int_{0}^{1} \frac{t + 1}{{x t}^{2} + t + 1} d t l e t d e c o m p o s e F (t) = \frac{t + 1}{{x t}^{2} + t + 1}$ $Δ = 1 - 4 x$ $c a s e 1 1 - 4 x > 0 \Leftrightarrow x < \frac{1}{4} \Rightarrow t_{1} = \frac{- 1 + \sqrt{1 - 4 x}}{2 x} a n d t_{2} = \frac{- 1 - \sqrt{1 - 4 x}}{2 x}$ $F (t) = \frac{t + 1}{x (t - t_{1}) (t - t_{2})} = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}}$ $a = \frac{t_{1} + 1}{x \frac{2 \sqrt{1 - 4 x}}{2 x}} = \frac{t_{1} + 1}{\sqrt{1 - 4 x}}$ $b = \frac{t_{2} + 1}{x (- \frac{2 \sqrt{1 - 4 x}}{2 x})} = - \frac{t_{2} + 1}{\sqrt{1 - 4 x}} \Rightarrow \int_{0}^{1} F (t) d t = \int_{0}^{1} {\frac{a}{t - t_{1}} + \frac{b}{t - t_{2}}} d t$ $= {[a l n ∣ t - t_{1} ∣ + b l n ∣ t - t_{2} ∣]}_{0}^{1} = a l n ∣ 1 - t_{1} ∣ + b l n ∣ 1 - t_{2} ∣ - a l n ∣ t_{1} ∣ - b l n ∣ t_{2} ∣$ $= a l n ∣ \frac{1 - t_{1}}{t_{1}} ∣ + b l n ∣ \frac{1 - t_{2}}{t_{2}} ∣ \Rightarrow$ $f^{^{'}} (x) = \frac{1}{x} - \frac{1}{x} {a l n ∣ \frac{1 - t_{1}}{t_{1}} ∣ + b l n ∣ \frac{1 - t_{2}}{t_{2}} ∣} \Rightarrow$ $f (x) = (1 - λ_{0}) l n ∣ x ∣ + c w i t h λ_{0} = a l n ∣ \frac{1 - t_{1}}{t_{1}} ∣ + b l n ∣ \frac{1 - t_{2}}{t_{2}} ∣ (k n o w n)$ $c = f (1) \Rightarrow f (x) = (1 - λ_{0}) l n ∣ x ∣ + f (1)$
Commented by maxmathsup by imad last updated on 11/Aug/18
$case 2 1−4x<0 ⇒no real roots if x>(1/4) ∫_0 ^1 F(x)dx = ∫_0 ^1 ((t+1)/(xt^2 +t +1)) dt =(1/(2x)) ∫_0 ^1 ((2xt +1 +2x −1)/(xt^2 +t +1)) dt =(1/(2x))[ln∣xt^2 +t+1∣]_0 ^1 +(1/(2x)) ∫_0 ^1 ((2x−1)/(xt^2 +t +1))dt =(1/(2x))ln∣ x+2∣ +((2x−1)/(2x^2 )) ∫_0 ^1 (dt/(t^2 +(t/x) +(1/x))) but ∫_0 ^1 (dt/(t^2 +(t/x)+(1/x))) = ∫_0 ^1 (dt/(t^2 +2(t/(2x)) + (1/(4x^2 )) +(1/x)−(1/(4x^2 )))) =∫_0 ^1 (dt/((t+(1/(2x)))^2 +((4x−1)/(4x^2 )))) =_(t+(1/(2x))= ((√(4x−1))/(2x))u) ∫_(1/(√(4x−1))) ^((2x+1)/(√(4x−1))) (1/(((4x−1)/(4x^2 ))(1+u^2 ))) ((√(4x−1))/(2x)) du =((4x^2 )/(4x−1)) .((√(4x−1))/(2x)) { arctan(((2x+1)/(√(4x−1))))−arctan((1/(√(4x−1))))} = ((2x)/(√(4x−1))) { arctan(((2x+1)/(√(4x−1))))−arctan((1/(√(4x−1))))}⇒ f^′ (x) = (1/x) −(2/(√(4x−1))){ arctan(((2x+1)/(√(4x−1))))−arctan((1/(√(4x−1))))} ⇒$
$c a s e 2 1 - 4 x < 0 \Rightarrow n o r e a l r o o t s i f x > \frac{1}{4}$ $\int_{0}^{1} F (x) d x = \int_{0}^{1} \frac{t + 1}{{x t}^{2} + t + 1} d t = \frac{1}{2 x} \int_{0}^{1} \frac{2 x t + 1 + 2 x - 1}{{x t}^{2} + t + 1} d t$ $= \frac{1}{2 x} {[l n ∣ {x t}^{2} + t + 1 ∣]}_{0}^{1} + \frac{1}{2 x} \int_{0}^{1} \frac{2 x - 1}{{x t}^{2} + t + 1} d t$ $= \frac{1}{2 x} l n ∣ x + 2 ∣ + \frac{2 x - 1}{2 x^{2}} \int_{0}^{1} \frac{d t}{t^{2} + \frac{t}{x} + \frac{1}{x}} b u t$ $\int_{0}^{1} \frac{d t}{t^{2} + \frac{t}{x} + \frac{1}{x}} = \int_{0}^{1} \frac{d t}{t^{2} + 2 \frac{t}{2 x} + \frac{1}{4 x^{2}} + \frac{1}{x} - \frac{1}{4 x^{2}}} = \int_{0}^{1} \frac{d t}{{(t + \frac{1}{2 x})}^{2} + \frac{4 x - 1}{4 x^{2}}}$ $=_{t + \frac{1}{2 x} = \frac{\sqrt{4 x - 1}}{2 x} u} \int_{\frac{1}{\sqrt{4 x - 1}}}^{\frac{2 x + 1}{\sqrt{4 x - 1}}} \frac{1}{\frac{4 x - 1}{4 x^{2}} (1 + u^{2})} \frac{\sqrt{4 x - 1}}{2 x} d u$ $= \frac{4 x^{2}}{4 x - 1} . \frac{\sqrt{4 x - 1}}{2 x} {a r c t a n (\frac{2 x + 1}{\sqrt{4 x - 1}}) - a r c t a n (\frac{1}{\sqrt{4 x - 1}})}$ $= \frac{2 x}{\sqrt{4 x - 1}} {a r c t a n (\frac{2 x + 1}{\sqrt{4 x - 1}}) - a r c t a n (\frac{1}{\sqrt{4 x - 1}})} \Rightarrow$ $f^{^{'}} (x) = \frac{1}{x} - \frac{2}{\sqrt{4 x - 1}} {a r c t a n (\frac{2 x + 1}{\sqrt{4 x - 1}}) - a r c t a n (\frac{1}{\sqrt{4 x - 1}})} \Rightarrow$
Commented by maxmathsup by imad last updated on 11/Aug/18

$f (x) = l n (x) - \int \frac{2 {a r c t a n (\frac{2 x + 1}{\sqrt{4 x - 1}}) - a r c t a n (\frac{1}{\sqrt{4 x - 1}})}}{\sqrt{4 x - 1}} d x + c . . b e$ $c o n t i n u e d . .$
Commented by maxmathsup by imad last updated on 11/Aug/18
$2) by parts ∫_0 ^1 ln(t^2 +t+1)dt =[tln(t^2 +t+1)]_0 ^1 −∫_0 ^1 t ((2t+1)/(t^2 +t+1))dt =ln(3) −∫_0 ^1 ((2t^2 +t)/(t^2 +t +1)) dt but ∫_0 ^1 ((2t^2 +t)/(t^2 +t +1)) dt = ∫_0 ^1 ((2(t^2 +t+1) +t−2t−2)/(t^2 +t+1)) dt=2 −∫_0 ^1 ((t+2)/(t^2 +t +1))dt =2−(1/2) ∫_0 ^1 ((2t+1+3)/(t^2 +t +1)) dt =2−(1/2)[ln∣t^2 +t+1∣_0 ^1 −(3/2) ∫_0 ^1 (dt/(t^2 +t+1)) =2−(1/2)ln(3) −(3/2) ∫_0 ^1 (dt/((t+(1/2))^2 +(3/4))) =_(t+(1/2)=((√3)/2)u) 2−((ln(3))/2) −(3/2) (4/3) ∫_(1/(√3)) ^(√3) (1/(1+u^2 )) ((√3)/2)du =2−((ln(3))/2) −(√3)[arctanu]_(1/(√3)) ^(√3) =2−((ln(3))/2) −(√3){ (π/3) −(π/6)} =2−((ln(3))/2) −(π/6)(√3) ⇒ ∫_0 ^1 ln(1+t +t^2 )dt =ln(3) −2 +((ln(3))/2) +((π(√3))/6) =(3/2)ln(3) +((π(√3))/6) −2 .$
$2) b y p a r t s \int_{0}^{1} l n (t^{2} + t + 1) d t = {[t l n (t^{2} + t + 1)]}_{0}^{1} - \int_{0}^{1} t \frac{2 t + 1}{t^{2} + t + 1} d t$ $= l n (3) - \int_{0}^{1} \frac{2 t^{2} + t}{t^{2} + t + 1} d t b u t$ $\int_{0}^{1} \frac{2 t^{2} + t}{t^{2} + t + 1} d t = \int_{0}^{1} \frac{2 (t^{2} + t + 1) + t - 2 t - 2}{t^{2} + t + 1} d t = 2 - \int_{0}^{1} \frac{t + 2}{t^{2} + t + 1} d t$ $= 2 - \frac{1}{2} \int_{0}^{1} \frac{2 t + 1 + 3}{t^{2} + t + 1} d t = 2 - \frac{1}{2} [l n ∣ t^{2} + t + 1 ∣_{0}^{1} - \frac{3}{2} \int_{0}^{1} \frac{d t}{t^{2} + t + 1}$ $= 2 - \frac{1}{2} l n (3) - \frac{3}{2} \int_{0}^{1} \frac{d t}{{(t + \frac{1}{2})}^{2} + \frac{3}{4}} =_{t + \frac{1}{2} = \frac{\sqrt{3}}{2} u} 2 - \frac{l n (3)}{2} - \frac{3}{2} \frac{4}{3} \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{1 + u^{2}} \frac{\sqrt{3}}{2} d u$ $= 2 - \frac{l n (3)}{2} - \sqrt{3} {[a r c t a n u]}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} = 2 - \frac{l n (3)}{2} - \sqrt{3} {\frac{π}{3} - \frac{π}{6}}$ $= 2 - \frac{l n (3)}{2} - \frac{π}{6} \sqrt{3} \Rightarrow$ $\int_{0}^{1} l n (1 + t + t^{2}) d t = l n (3) - 2 + \frac{l n (3)}{2} + \frac{π \sqrt{3}}{6} = \frac{3}{2} l n (3) + \frac{π \sqrt{3}}{6} - 2 .$
Commented by maxmathsup by imad last updated on 11/Aug/18

$3) \int_{0}^{1} l n (1 - t^{3}) d t = \int_{0}^{1} l n (1 - t) d t + \int_{0}^{1} l n (1 + t + t^{2}) d t b u t$ $\int_{0}^{1} l n (1 - t) d t =_{1 - t = x} \int_{0}^{1} l n (x) d x = {[x l n (x) - x]}_{0}^{1} = - 1 \Rightarrow$ $\int_{0}^{1} l n (1 - x^{3}) d x = - 1 + \frac{3}{2} l n (3) + \frac{π \sqrt{3}}{6} - 2$ $= - 3 + \frac{3}{2} l n (3) + \frac{π \sqrt{3}}{6} .$
	Answered by alex041103 last updated on 11/Aug/18

	$1 + t + {x t}^{2} = x (t - t_{1}) (t - t_{2})$ $t_{1, 2} = \frac{- 1 \pm \sqrt{1 - 4 x}}{2 x}$ $\Rightarrow f (x) = \int_{0}^{1} l n (x) d t + \int_{0}^{1} l n (t - t_{1}) d t + \int_{0}^{1} l n (t - t_{2}) d t$ $A s w e k n o w \int l n (x) d x = x (l n (x) - 1)$ $\Rightarrow f (x) = l n (x) + l n (1 - t_{1}) + l n (1 - t_{2}) - 2 =$ $= l n (x (1 - t_{1}) (1 - t_{2})) - 2 =$ $= l n (2 + x) - 2$ $\Rightarrow f (x) = l n (2 + x) - 2$ $\Rightarrow \int_{0}^{1} l n (1 + t + t^{2}) d t = f (x = 1) = l n (3) - 2$ $\Rightarrow \int_{0}^{1} l n (1 - t^{3}) d t$ $1 - t^{3} = (1 - t) (1 + t + t^{2})$ $\Rightarrow \int_{0}^{1} l n (1 - t^{3}) d t = \int_{0}^{1} l n (1 - t) d t + f (1) =$ $= {[(1 - t) (l n (1 - t) - 1)]}_{1}^{0} + f (1) =$ $= [- 1 - \underset{t \to 1^{-}}{l i m} \frac{l n (1 - t) - 1}{\frac{1}{1 - t}}] + f (1) =$ $= - [1 + \underset{t \to 1^{-}}{l i m} \frac{- \frac{1}{1 - t}}{\frac{1}{{(1 - t)}^{2}}}] + f (1) =$ $= - 1 + l n (3) - 2$ $\Rightarrow \int_{0}^{1} l n (1 - t^{3}) d t = l n (3) - 3$
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