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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) 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) \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) = \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 .