decompose-F-x-2x-1-x-2-1-2-x-2-3-and-calculate-2-F-x-dx-

Question Number 119425 by mathmax by abdo last updated on 24/Oct/20

decompose F (x) = \frac{2 x - 1}{{(x^{2} - 1)}^{2} (x^{2} + 3)}

and calculate \int_{\sqrt{2}}^{+ \infty} F (x) dx

Answered by 1549442205PVT last updated on 24/Oct/20

F (x) = \frac{2 x - 1}{{(x^{2} - 1)}^{2} (x^{2} + 3)} = \frac{ax + b}{x^{2} + 3} + \frac{{mx}^{3} + {nx}^{2} + px + q}{x^{4} - {2 x}^{2} + 1}

\Leftrightarrow (a + m) x^{5} + (b + n) x^{4} + (3 m + p - 2 a) x^{3}

+ (3 n + q - 2 b) x^{2} + (a + 3 p) x + b + 3 q = 2 x - 1

\Leftrightarrow {\begin{cases} a + m = 0 \\ b + n = 0 \\ 3 m + p - 2 a = 0 \\ 3 n + q - 2 b = 0 \\ a + 3 p = 2 \\ b + 3 q = - 1 \end{cases} {\begin{cases} 5 m + p = 0 \\ 5 n + q = 0 \\ - m + 3 p = 2 \\ - n + 3 q = - 1 \end{cases}

16 q = - 5 \Rightarrow q = - 5 / 16 \Rightarrow n = 1 / 16

16 p = 10 \Rightarrow p = 10 / 16 \Rightarrow m = - 2 / 16

\Rightarrow a = 2 / 16, b = - 1 / 16

Hence we have

F (x) = \frac{2 x - 1}{{(x^{2} - 1)}^{2} (x^{2} + 3)}

= \frac{2 x - 1}{16 (x^{2} + 3)} - \frac{{2 x}^{3} - x^{2} - 10 x + 5}{16 {(x^{2} - 1)}^{2}}

{2 x}^{3} - x^{2} - 10 x + 5 = (x^{2} - 1) (2 x - 1) - 8 x + 4

\Rightarrow \frac{{2 x}^{3} - x^{2} - 10 x + 5}{16 {(x^{2} - 1)}^{2}} = \frac{2 x - 1}{16 (x^{2} - 1)} - \frac{8 x - 4}{16 {(x^{2} - 1)}^{2}}

\frac{2}{x^{2} - 1} = \frac{1}{x - 1} - \frac{1}{x + 1} = a - b \Rightarrow 2 ab = a - b

\Rightarrow \frac{4}{{(x^{2} - 1)}^{2}} = {(a - b)}^{2} = a^{2} + b^{2} - (a - b)

\Rightarrow F (x) = \frac{2 x - 1}{16 (x^{2} + 3)} - \frac{2 x - 1}{16 (x^{2} - 1)} + \frac{4 (2 x - 1)}{16 {(x^{2} - 1)}^{2}}

\int_{\sqrt{2}}^{\infty} F (x) = \int_{\sqrt{2}}^{\infty} \frac{d (x^{2} + 3)}{16 (x^{2} + 3} - \frac{1}{16} \int_{\sqrt{2}}^{\infty} \frac{dx}{(x^{2} + 3)}

- \frac{1}{16} \int_{\sqrt{2}}^{\infty} \frac{d (x^{2} - 1)}{x^{2} - 1} + \frac{1}{16} \int_{\sqrt{2}}^{\infty} \frac{dx}{x^{2} - 1} + \frac{1}{4} \int_{\sqrt{2}}^{\infty} \frac{d (x^{2} - 1)}{{(x^{2} - 1)}^{2}}

- \frac{1}{4} \int_{\sqrt{2}}^{\infty} \frac{dx}{{(x^{2} - 1)}^{2}} (1) . We have

J = \frac{1}{16} \int_{\sqrt{2}}^{\infty} \frac{dx}{(x^{2} - 1)} = \frac{1}{32} \int_{\sqrt{2}}^{\infty} (\frac{1}{x - 1} - \frac{1}{x + 1}) dx

= \frac{1}{32} \ln ∣ \frac{x - 1}{x + 1} ∣_{\sqrt{2}}^{\infty} = - \frac{1}{32} \ln \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = - \frac{1}{16} \ln (\sqrt{2} - 1)

I = \frac{1}{4} \int_{\sqrt{2}}^{\infty} \frac{dx}{{(x^{2} - 1)}^{2}} = \frac{1}{16} \int_{\sqrt{2}}^{\infty} {(\frac{1}{x - 1} - \frac{1}{x + 1})}^{2} dx =

\frac{1}{16} \int_{\sqrt{2}}^{\infty} (\frac{1}{{(x - 1)}^{2}} + \frac{1}{{(x + 1)}^{2}} - \frac{1}{x - 1} + \frac{1}{x + 1}) dx (2)

= \frac{1}{16} [{[- \frac{1}{x - 1} - \frac{1}{x + 1} + \ln ∣ \frac{x + 1}{x - 1 ∣}]}_{\sqrt{2}}^{\infty}

= \frac{1}{64} (\frac{1}{\sqrt{2} - 1} + \frac{1}{\sqrt{2} + 1} - \ln \frac{\sqrt{2} + 1}{\sqrt{2} - 1}) = \frac{1}{16} [2 \sqrt{2} - 2 \ln (\sqrt{2} + 1)]

From (1) (2) we get

\int_{\sqrt{2}}^{\infty} F (x) dx = \int_{\sqrt{2}}^{\infty} \frac{2 x - 1}{{(x^{2} - 1)}^{2} (x^{2} + 3)} dx

= [\frac{1}{16} \ln (x^{2} + 3) - \frac{1}{16} . \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}})

{- \frac{1}{16} \ln ∣ x^{2} - 1 ∣ - \frac{1}{4 (x^{2} - 1)}]}_{\sqrt{2}}^{\infty} + J - I

{= \frac{1}{16} \ln \frac{x^{2} + 3}{x^{2} - 1} - \frac{1}{16 \sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - \frac{1}{4 (x^{2} - 1)}]}_{\sqrt{2}}^{\infty}

- \frac{1}{16} \ln (\sqrt{2} - 1) - \frac{1}{16} (2 \sqrt{2} - 2 \ln (1 + \sqrt{2})

= - \frac{π}{32 \sqrt{3}} - \frac{1}{16} \ln 5 + \frac{1}{16 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{6}}{3}) + \frac{1}{4}

- \frac{\sqrt{2}}{8} + \frac{3}{16} \ln (\sqrt{2} + 1) \approx 0.1059174283

Commented by Bird last updated on 24/Oct/20

t h a n k y o u s i r

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