calculate ∫_1 ^(3 ) ((x−2)/(√(x^2 +x+1)))dx


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Question Number 73181 by mathmax by abdo last updated on 07/Nov/19

$c a l c u l a t e \int_{1}^{3} \frac{x - 2}{\sqrt{x^{2} + x + 1}} d x$
Commented by mathmax by abdo last updated on 07/Nov/19
$we have x^2 +x+1 =(x+(1/2))^2 +(3/4) changement x+(1/2)=((√3)/2)sh(t)give sh(t)=((2x+1)/(√3)) ⇒ ∫_1 ^3 ((x−2)/(√(x^2 +x+1)))dx =∫_(arsh((√3))) ^(argsh((7/(√3)))) ((((√3)/2)sh(t)−(1/2)−2)/(((√3)/2)ch(t)))×((√3)/2)ch(t)dt =∫_(ln((√3)+(√4))) ^(ln((7/(√3))+(√(1+((49)/3) )))) (((√3)/2)sh(t)−(5/2))dt =((√3)/2) ∫_(ln((√3)+2)) ^(ln((7/((√3) ))+(√(1+((49)/3))))) sh(t)dt−(5/2)(ln((7/(√3))+((√(52))/(√3)))−ln(2+(√3))) =((√3)/4)[ e^t +e^(−t) ]_(ln(2+(√3))) ^(ln(((7+(√(52)))/(√3)))) −(5/2){ln(((7+(√(52)))/(√3)))−ln(2+(√3))} =((√3)/4){(((7+(√(52)))/(√3)))+(((7+(√(52)))/(√3)))^(−1) −(2+(√3))−(2+(√3))^(−1) } −(5/2){ln(((7+(√(52)))/(√3)))−ln(2+(√3))}$
$w e h a v e x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4} c h a n g e m e n t x + \frac{1}{2} = \frac{\sqrt{3}}{2} s h (t) g i v e$ $s h (t) = \frac{2 x + 1}{\sqrt{3}} \Rightarrow$ $\int_{1}^{3} \frac{x - 2}{\sqrt{x^{2} + x + 1}} d x = \int_{a r s h (\sqrt{3})}^{a r g s h (\frac{7}{\sqrt{3}})} \frac{\frac{\sqrt{3}}{2} s h (t) - \frac{1}{2} - 2}{\frac{\sqrt{3}}{2} c h (t)} \times \frac{\sqrt{3}}{2} c h (t) d t$ $= \int_{l n (\sqrt{3} + \sqrt{4})}^{l n (\frac{7}{\sqrt{3}} + \sqrt{1 + \frac{49}{3}})} (\frac{\sqrt{3}}{2} s h (t) - \frac{5}{2}) d t$ $= \frac{\sqrt{3}}{2} \int_{l n (\sqrt{3} + 2)}^{l n (\frac{7}{\sqrt{3}} + \sqrt{1 + \frac{49}{3}})} s h (t) d t - \frac{5}{2} (l n (\frac{7}{\sqrt{3}} + \frac{\sqrt{52}}{\sqrt{3}}) - l n (2 + \sqrt{3}))$ $= \frac{\sqrt{3}}{4} {[e^{t} + e^{- t}]}_{l n (2 + \sqrt{3})}^{l n (\frac{7 + \sqrt{52}}{\sqrt{3}})} - \frac{5}{2} {l n (\frac{7 + \sqrt{52}}{\sqrt{3}}) - l n (2 + \sqrt{3})}$ $= \frac{\sqrt{3}}{4} {(\frac{7 + \sqrt{52}}{\sqrt{3}}) + {(\frac{7 + \sqrt{52}}{\sqrt{3}})}^{- 1} - (2 + \sqrt{3}) - {(2 + \sqrt{3})}^{- 1}}$ $- \frac{5}{2} {l n (\frac{7 + \sqrt{52}}{\sqrt{3}}) - l n (2 + \sqrt{3})}$
	Answered by petrochengula last updated on 07/Nov/19

	$c o n s i d e r I = \int \frac{x - 2}{\sqrt{x^{2} + x + 1}} d x$ $= \frac{1}{2} \int \frac{2 (x - 2)}{\sqrt{x^{2} + x + 1}} d x = \frac{1}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x + 1}} d x - \frac{1}{2} \int \frac{5}{\sqrt{x^{2} + x + 1}} d x$ $= \sqrt{x^{2} + x + 1} - \frac{5}{2} \int \frac{1}{\sqrt{x^{2} + x + \frac{1}{4} + \frac{3}{4}}} d x$ $= \sqrt{x^{2} + x + 1} - \frac{5}{2} \int \frac{1}{\sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}}$ $c o n s i d e r \int \frac{1}{\sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}} d x$ $= \int \frac{1}{\sqrt{\frac{3}{4} (\frac{4}{3} {(\frac{2 x + 1}{2})}^{2} + 1)}} d x$ $= \frac{2}{\sqrt{3}} \int \frac{1}{\sqrt{{(\frac{2 x + 1}{\sqrt{3}})}^{2} + 1}} d x$ $l e t s i n h θ = \frac{2 x + 1}{\sqrt{3}} \Rightarrow \sqrt{3} s i n h θ = 2 x + 1 \Rightarrow \sqrt{3} c o s h θ d θ = 2 \frac{d x}{d θ} \Rightarrow d x = \frac{\sqrt{3}}{2} c o s h θ d θ = d x$ $= \int d θ$ $= θ + C$ $= {s i n h}^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + c$ $\int \frac{x - 2}{\sqrt{x^{2} + x + 1}} d x = \sqrt{x^{2} + x + 1} - \frac{5}{2} {s i n h}^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + C$ $\int_{1}^{3} \frac{x - 2}{\sqrt{x^{2} + x + 1}} d x = \sqrt{13} - \frac{5}{2} {s i n h}^{- 1} (\frac{7}{\sqrt{3}}) - \sqrt{3} + \frac{5}{2} {s i n h}^{- 1} (\frac{3}{\sqrt{3}})$
	Commented by petrochengula last updated on 07/Nov/19

	$p l e a s e c h e c k i f i t i s c o r r e c t$
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