1) calculate ∫ (x^2 −x+2)(√(x^2 −x+1))dx 2)find the value of ∫


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Question Number 63722 by mathmax by abdo last updated on 08/Jul/19

$1) c a l c u l a t e \int (x^{2} - x + 2) \sqrt{x^{2} - x + 1} d x$ $2) f i n d t h e v a l u e o f \int_{0}^{1} (x^{2} - x + 2) \sqrt{x^{2} - x + 1} d x .$
Commented by Prithwish sen last updated on 08/Jul/19

$1) putting x^{2} - x + 1 = u^{2}$ $\Rightarrow (2 x - 1) dx = 2 udu and (2 x - 1) = \pm \sqrt{4 u - 3 .}$ $= \pm 2 \int \frac{u^{2} (u + 1) du}{\sqrt{4 u - 3}} putting 4 u - 3 = t^{2} \Rightarrow 2 du = tdt$ $= \frac{1}{64} \int {(t^{2} + 3)}^{2} (t^{2} + 7) dt$ $= \frac{1}{448} {(4 \sqrt{x^{2} - x + 1} - 3)}^{\frac{7}{2}} + \frac{13}{320} {(4 \sqrt{x^{2} - x + 1} - 3)}^{\frac{5}{2}} + \frac{15}{192} {(4 \sqrt{x^{2} - x + 1} - 3)}^{\frac{3}{2}} + \frac{63}{64} (4 \sqrt{x^{2} - x + 1} - 3) + C$ $please check .$
Commented by MJS last updated on 08/Jul/19

$something went wrong$ $you can put t = \sqrt{x^{2} - x + 1} \to d x = 2 d t \frac{\sqrt{x^{2} - x + 1}}{2 x - 1}; x = \frac{1 + \sqrt{4 t^{2} - 3}}{}$ $\Rightarrow 2 \int \frac{t^{2} (t^{2} + 1)}{\sqrt{4 t^{2} - 3}}$ $or you can put t = x^{2} - x + 1 \to d x = \frac{d t}{2 x - 1}; x = \frac{1 + \sqrt{4 u - 3}}{2}$ $\Rightarrow \int \frac{\sqrt{t} (t + 1)}{\sqrt{4 t - 3}}$
Commented by Prithwish sen last updated on 08/Jul/19

$thank you sir . I will check it .$
Commented by mathmax by abdo last updated on 08/Jul/19
$1) let I =∫ (x^2 −x+2)(√(x^2 −x+1))dx we have x^2 −x +1 =x^2 −2(1/2)x +(1/4) +1−(1/4) =(x−(1/2))^2 +(3/4) so we use the changement x−(1/2) =((√3)/2)sh(t) ⇒ ⇒ I = ∫ ( ((1/2)+((√3)/2)sh(t))^2 −((1/2)+((√3)/2)sh(t)) +2)((√3)/2)ch(t)dt =((√3)/2)∫ { (1/4)(1+2(√3)sh(t) +3sh^2 (t)−((√3)/2)sh(t)+2−(1/2))ch(t)dt =((√3)/2) ∫ { ((√3)/2)sh(t) +(3/4)sh^2 (t)−((√3)/2)sh(t) +(7/4)}ch(t)dt =((3(√3))/8) ∫ sh(t)ch(t)dt +((7(√3))/8) ∫ ch(t)dt =((3(√3))/(16))sh^2 (t) +((7(√3))/8) sh(t) +c =((3(√3))/(16))((ch(2t)−1)/2) +((7(√3))/8)sh(t) +c =((3(√3))/(32)){ ch(2t)−1} +((7(√3))/8)sh(t) +c . we have t =argsh(((2x−1)/(√3))) =ln(((2x−1)/(√3)) +(√(1+(((2x−1)/(√3)))^2 ))) ch(2t) =((e^(2t) +e^(−2t) )/2) =(1/2){(((2x−1)/(√3)) +(√(1+(((2x−1)/(√3)))^2 )))^2 +(1/((((2x−1)/(√3))+(√(1+(((2x−1)/((√3)))))^2 ))))} sh(t) =((e^t −e^(−t) )/2) =(1/2){ ((2x−1)/(√3))+(√(1+(((2x−1)/(√3)))))^2 − (1/(((2x−1)/(√3))+(√(1+(((2x−1)/(√3)))))^2 ))} so the value of I is known.$
$1) l e t I = \int (x^{2} - x + 2) \sqrt{x^{2} - x + 1} d x w e h a v e$ $x^{2} - x + 1 = x^{2} - 2 \frac{1}{2} x + \frac{1}{4} + 1 - \frac{1}{4} = {(x - \frac{1}{2})}^{2} + \frac{3}{4} s o w e u s e t h e$ $c h a n g e m e n t x - \frac{1}{2} = \frac{\sqrt{3}}{2} s h (t) \Rightarrow$ $\Rightarrow I = \int ({(\frac{1}{2} + \frac{\sqrt{3}}{2} s h (t))}^{2} - (\frac{1}{2} + \frac{\sqrt{3}}{2} s h (t)) + 2) \frac{\sqrt{3}}{2} c h (t) d t$ $= \frac{\sqrt{3}}{2} \int {\frac{1}{4} (1 + 2 \sqrt{3} s h (t) + 3 {s h}^{2} (t) - \frac{\sqrt{3}}{2} s h (t) + 2 - \frac{1}{2}) c h (t) d t$ $= \frac{\sqrt{3}}{2} \int {\frac{\sqrt{3}}{2} s h (t) + \frac{3}{4} {s h}^{2} (t) - \frac{\sqrt{3}}{2} s h (t) + \frac{7}{4}} c h (t) d t$ $= \frac{3 \sqrt{3}}{8} \int s h (t) c h (t) d t + \frac{7 \sqrt{3}}{8} \int c h (t) d t$ $= \frac{3 \sqrt{3}}{16} {s h}^{2} (t) + \frac{7 \sqrt{3}}{8} s h (t) + c$ $= \frac{3 \sqrt{3}}{16} \frac{c h (2 t) - 1}{2} + \frac{7 \sqrt{3}}{8} s h (t) + c$ $= \frac{3 \sqrt{3}}{32} {c h (2 t) - 1} + \frac{7 \sqrt{3}}{8} s h (t) + c .$ $w e h a v e t = a r g s h (\frac{2 x - 1}{\sqrt{3}}) = l n (\frac{2 x - 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x - 1}{\sqrt{3}})}^{2}})$ $c h (2 t) = \frac{e^{2 t} + e^{- 2 t}}{2} = \frac{1}{2} {{(\frac{2 x - 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x - 1}{\sqrt{3}})}^{2}})}^{2} + \frac{1}{(\frac{2 x - 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x - 1}{\sqrt{3})})}^{2}}}}$ $s h (t) = \frac{e^{t} - e^{- t}}{2} = \frac{1}{2} {{\frac{2 x - 1}{\sqrt{3}} + \sqrt{1 + (\frac{2 x - 1}{\sqrt{3}}})}^{2} - \frac{1}{{\frac{2 x - 1}{\sqrt{3}} + \sqrt{1 + (\frac{2 x - 1}{\sqrt{3}}})}^{2}}}$ $s o t h e v a l u e o f I i s k n o w n .$
Commented by mathmax by abdo last updated on 08/Jul/19
$2) ∫_0 ^1 (x^2 −x+2)(√(x^2 −x+1))dx =[F(x)]_0 ^1 =F(1)−F(0) F(1)=((3(√3))/(32)){(1/2)((1/(√3)) +(√(1+(1/3))))^2 +(1/(((1/(√3))+(√(1+(1/3))))^2 ))}−1} +((7(√3))/(16)){ (1/(√3)) +(√(1+(1/3))) −(1/((1/(√3))+(√(1+(1/3)))))} F(0) =((3(√3))/(32)){(−(1/3) +(√(1+(1/3))))^2 +(1/((−(1/3) +(√(1+(1/3))))^2 ))} +((7(√3))/(16)){−(1/(√3)) +(√(1+(1/3)))−(1/(−(1/(√3))+(√(1+(1/3)))))}$
$2) \int_{0}^{1} (x^{2} - x + 2) \sqrt{x^{2} - x + 1} d x = {[F (x)]}_{0}^{1} = F (1) - F (0)$ $F (1) = \frac{3 \sqrt{3}}{32} {\frac{1}{2} {(\frac{1}{\sqrt{3}} + \sqrt{1 + \frac{1}{3}})}^{2} + \frac{1}{{(\frac{1}{\sqrt{3}} + \sqrt{1 + \frac{1}{3}})}^{2}}} - 1}$ $+ \frac{7 \sqrt{3}}{16} {\frac{1}{\sqrt{3}} + \sqrt{1 + \frac{1}{3}} - \frac{1}{\frac{1}{\sqrt{3}} + \sqrt{1 + \frac{1}{3}}}}$ $F (0) = \frac{3 \sqrt{3}}{32} {{(- \frac{1}{3} + \sqrt{1 + \frac{1}{3}})}^{2} + \frac{1}{{(- \frac{1}{3} + \sqrt{1 + \frac{1}{3}})}^{2}}}$ $+ \frac{7 \sqrt{3}}{16} {- \frac{1}{\sqrt{3}} + \sqrt{1 + \frac{1}{3}} - \frac{1}{- \frac{1}{\sqrt{3}} + \sqrt{1 + \frac{1}{3}}}}$
Commented by mathmax by abdo last updated on 08/Jul/19
$F(0) =((3(√3))/(32)){ (−(1/(√3)) +(2/(√3)))^2 +(1/((−(1/(√3))+(2/(√3)))^2 ))}+((7(√3))/(16)){−(1/(√3)) +(2/(√3)) −(1/(−(1/(√3))+(2/(√3))))} =((3(√3))/(32)){ (1/3) +3}+((7(√3))/(16)){ (1/(√3)) −(√3)}$
$F (0) = \frac{3 \sqrt{3}}{32} {{(- \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})}^{2} + \frac{1}{{(- \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})}^{2}}} + \frac{7 \sqrt{3}}{16} {- \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}$ $- \frac{1}{- \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}}} = \frac{3 \sqrt{3}}{32} {\frac{1}{3} + 3} + \frac{7 \sqrt{3}}{16} {\frac{1}{\sqrt{3}} - \sqrt{3}}$
	Answered by MJS last updated on 08/Jul/19

	$\int (x^{2} - x + 2) \sqrt{x^{2} - x + 1} d x =$ $[t = \sinh^{- 1} \frac{\sqrt{3} (2 x - 1)}{3} \to d x = \sqrt{x^{2} - x + 1} d t]$ $= \frac{75}{128} \int d t + \frac{9}{128} \int \cosh 4 t d t + \frac{21}{32} \int \cosh 2 t d t =$ $= \frac{75}{128} t + \frac{9}{512} \sinh 4 t + \frac{21}{64} \sinh 2 t =$ $= \frac{75}{128} \sinh^{- 1} \frac{\sqrt{3} (2 x - 1)}{3} + \frac{1}{64} (2 x - 1) (8 x^{2} - 8 x + 5) \sqrt{x^{2} - x + 1} + \frac{7}{16} (2 x - 1) \sqrt{x^{2} - x + 1} =$ $= \frac{75}{128} \sinh^{- 1} \frac{\sqrt{3} (2 x - 1)}{3} + \frac{1}{64} (2 x - 1) (8 x^{2} - 8 x + 33) \sqrt{x^{2} - x + 1} + C$ $\underset{0}{\int^{1}} (x^{2} - x + 2) \sqrt{x^{2} - x + 1} d x = \frac{33}{32} + \frac{75}{64} \sinh^{- 1} \frac{\sqrt{3}}{3} = \frac{33}{32} + \frac{75}{128} \ln 3$
	Commented by Prithwish sen last updated on 08/Jul/19

	$sir please check mine . I have just try it$ $another way .$
	Commented by mathmax by abdo last updated on 08/Jul/19

	$t h a n k y o u s i r m j s .$
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