eplcit f(x)=∫_0 ^1 ln(x+t+t^2 )dt with x>(1/4) 2)calculate ∫


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Question Number 73335 by mathmax by abdo last updated on 10/Nov/19

$e p l c i t f (x) = \int_{0}^{1} l n (x + t + t^{2}) d t w i t h x > \frac{1}{4}$ $2) c a l c u l a t e \int_{0}^{1} l n (t^{2} + t + \sqrt{2}) d t$
Commented by mathmax by abdo last updated on 10/Nov/19
$1) we have f(x)=∫_0 ^1 ln(t^2 +t+x)dt by parts we get f(x)=[tln(t^2 +t+x)]_0 ^1 −∫_0 ^1 t×((2t+1)/(t^2 +t +x))dt =ln(2+x)−∫_0 ^1 ((2t^2 +t)/(t^2 +t+x))dt we have ∫_0 ^1 ((2t^2 +t)/(t^2 +t +x))dt =∫_0 ^1 ((2(t^2 +t+x)−2t−2x+t)/(t^2 +t+x))dt =∫_0 ^1 (2−((t+2x)/(t^2 +t+x)))dt =2 −∫_0 ^1 ((t+2x)/(t^2 +t+x))dt =2−(1/2)∫_0 ^1 ((2t +4x+1−1)/(t^(2 ) +t+x))dt =2−(1/2)∫_0 ^1 ((2t+1)/(t^2 +t+x))dt−(1/2)∫_0 ^1 ((4x−1)/(t^2 +t+x))dt =2−(1/2)[ln(t^2 +t+x)]_0 ^1 −((4x−1)/2)∫_0 ^1 (dt/(t^2 +t+x)) =2−(1/2)(ln(2+x)−ln(x))−((4x−1)/2)∫_0 ^1 (dt/(t^2 +t+x)) ∫_0 ^1 (dt/(t^2 +t+x)) =∫_0 ^1 (dt/(t^2 +2(t/2)+(1/4)+x−(1/4))) =∫_0 ^1 (dt/((t+(1/2))^2 +((4x−1)/4))) (4x−1>0) =_(t+(1/2)=((√(4x−1))/2)u) (4/(4x−1)) ∫_(1/(√(4x−1))) ^(3/(√(4x−1))) (1/(1+u^2 )) ((√(4x−1))/2) du =(2/(√(4x−1))){ arctan((3/(√(4x−1))))−arctan((1/(√(4x−1))))} ⇒ f(x)=ln(2+x)−2+(1/2)ln(((2+x)/x))+((4x−1)/2)×(2/(√(4x−1))){arctan((3/(√(4x−1))) −arctan((1/(√(4x−1))))} f(x)=(3/2)ln(2+x)−(1/2)ln(x)−2 +(√(4x−1)){arctan((3/(√(4x−1))))−arctan((1/(√(4x−1))))}$
$1) w e h a v e f (x) = \int_{0}^{1} l n (t^{2} + t + x) d t b y p a r t s w e g e t$ $f (x) = {[t l n (t^{2} + t + x)]}_{0}^{1} - \int_{0}^{1} t \times \frac{2 t + 1}{t^{2} + t + x} d t$ $= l n (2 + x) - \int_{0}^{1} \frac{2 t^{2} + t}{t^{2} + t + x} d t w e h a v e$ $\int_{0}^{1} \frac{2 t^{2} + t}{t^{2} + t + x} d t = \int_{0}^{1} \frac{2 (t^{2} + t + x) - 2 t - 2 x + t}{t^{2} + t + x} d t$ $= \int_{0}^{1} (2 - \frac{t + 2 x}{t^{2} + t + x}) d t = 2 - \int_{0}^{1} \frac{t + 2 x}{t^{2} + t + x} d t$ $= 2 - \frac{1}{2} \int_{0}^{1} \frac{2 t + 4 x + 1 - 1}{t^{2} + t + x} d t = 2 - \frac{1}{2} \int_{0}^{1} \frac{2 t + 1}{t^{2} + t + x} d t - \frac{1}{2} \int_{0}^{1} \frac{4 x - 1}{t^{2} + t + x} d t$ $= 2 - \frac{1}{2} {[l n (t^{2} + t + x)]}_{0}^{1} - \frac{4 x - 1}{2} \int_{0}^{1} \frac{d t}{t^{2} + t + x}$ $= 2 - \frac{1}{2} (l n (2 + x) - l n (x)) - \frac{4 x - 1}{2} \int_{0}^{1} \frac{d t}{t^{2} + t + x}$ $\int_{0}^{1} \frac{d t}{t^{2} + t + x} = \int_{0}^{1} \frac{d t}{t^{2} + 2 \frac{t}{2} + \frac{1}{4} + x - \frac{1}{4}} = \int_{0}^{1} \frac{d t}{{(t + \frac{1}{2})}^{2} + \frac{4 x - 1}{4}} (4 x - 1 > 0)$ $=_{t + \frac{1}{2} = \frac{\sqrt{4 x - 1}}{2} u} \frac{4}{4 x - 1} \int_{\frac{1}{\sqrt{4 x - 1}}}^{\frac{3}{\sqrt{4 x - 1}}} \frac{1}{1 + u^{2}} \frac{\sqrt{4 x - 1}}{2} d u$ $= \frac{2}{\sqrt{4 x - 1}} {a r c t a n (\frac{3}{\sqrt{4 x - 1}}) - a r c t a n (\frac{1}{\sqrt{4 x - 1}})} \Rightarrow$ $f (x) = l n (2 + x) - 2 + \frac{1}{2} l n (\frac{2 + x}{x}) + \frac{4 x - 1}{2} \times \frac{2}{\sqrt{4 x - 1}} {a r c t a n (\frac{3}{\sqrt{4 x - 1}}$ $- a r c t a n (\frac{1}{\sqrt{4 x - 1}})}$ $f (x) = \frac{3}{2} l n (2 + x) - \frac{1}{2} l n (x) - 2 + \sqrt{4 x - 1} {a r c t a n (\frac{3}{\sqrt{4 x - 1}}) - a r c t a n (\frac{1}{\sqrt{4 x - 1}})}$
Commented by mathmax by abdo last updated on 10/Nov/19
$2) ∫_0 ^1 ln(t^2 +t+(√2))dt =f((√2)) =(3/2)ln(2+(√2))−(1/4)ln(2)−2+(√(4(√2)−1)){ arctan((3/(√(4(√2)−1))))−arctan((1/(√(4(√2)−1))))}$
$2) \int_{0}^{1} l n (t^{2} + t + \sqrt{2}) d t = f (\sqrt{2})$ $= \frac{3}{2} l n (2 + \sqrt{2}) - \frac{1}{4} l n (2) - 2 + \sqrt{4 \sqrt{2} - 1} {a r c t a n (\frac{3}{\sqrt{4 \sqrt{2} - 1}}) - a r c t a n (\frac{1}{\sqrt{4 \sqrt{2} - 1}})}$
	Answered by mind is power last updated on 10/Nov/19
	$f(x)=∫_0 ^1 ln(x+t+t^2 )dt=ln(x+2)−ln(x)−∫(((1+2t)t)/(x+t+t^2 ))dt {by part} =ln(((x+2)/2))−∫_0 ^1 ((2t^2 +2t+2x−t−(1/2)+(1/2)−2x)/(t^2 +t+x))dt =ln(((x+2)/x))−2∫1dt+∫((t+(1/2))/(t^2 +t+x))+(2x−(1/2))∫_0 ^1 (dt/((t+(1/2))^2 +((4x−1)/4))) =ln(((x+2)/x))−2+(1/2)ln{(((x+2)/x))}+∫_0 ^1 (dt/((((2t)/(√(4x−1)))+(1/(√(4x−1))))^2 +1)) =(3/2)ln(((x+2)/x))−2+((√(4x−1))/2).[arctan(((2t+1)/(√(4x−1))))]_0 ^1 =(3/2)ln(((x+2)/x))−2+((√(4x−1))/2).[arctab((3/(√(4x−1))))−arctan((1/(√(4x−1))))] 2)x=(√2)$
	$f (x) = \int_{0}^{1} \ln (x + t + t^{2}) dt = \ln (x + 2) - \ln (x) - \int \frac{(1 + 2 t) t}{x + t + t^{2}} dt {by part}$ $= \ln (\frac{x + 2}{2}) - \int_{0}^{1} \frac{{2 t}^{2} + 2 t + 2 x - t - \frac{1}{2} + \frac{1}{2} - 2 x}{t^{2} + t + x} dt$ $= \ln (\frac{x + 2}{x}) - 2 \int 1 dt + \int \frac{t + \frac{1}{2}}{t^{2} + t + x} + (2 x - \frac{1}{2}) \int_{0}^{1} \frac{dt}{{(t + \frac{1}{2})}^{2} + \frac{4 x - 1}{4}}$ $= \ln (\frac{x + 2}{x}) - 2 + \frac{1}{2} \ln {(\frac{x + 2}{x})} + \int_{0}^{1} \frac{dt}{{(\frac{2 t}{\sqrt{4 x - 1}} + \frac{1}{\sqrt{4 x - 1}})}^{2} + 1}$ $= \frac{3}{2} \ln (\frac{x + 2}{x}) - 2 + \frac{\sqrt{4 x - 1}}{2} . {[\arctan (\frac{2 t + 1}{\sqrt{4 x - 1}})]}_{0}^{1}$ $= \frac{3}{2} \ln (\frac{x + 2}{x}) - 2 + \frac{\sqrt{4 x - 1}}{2} . [arctab (\frac{3}{\sqrt{4 x - 1}}) - \arctan (\frac{1}{\sqrt{4 x - 1}})]$ $2) x = \sqrt{2}$
	Commented by mathmax by abdo last updated on 10/Nov/19

	$t h a n k s s i r .$
	Commented by mind is power last updated on 10/Nov/19

	$y^{'} re welcom$
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