calculate-1-dx-2x-2-1-5-

Question Number 116245 by mathmax by abdo last updated on 02/Oct/20

calculate \int_{1}^{\infty} \frac{dx}{{({2 x}^{2} - 1)}^{5}}

Answered by MJS_new last updated on 02/Oct/20

\int \frac{d x}{{(2 x^{2} - 1)}^{5}} =

[Ostrogradski]

= \frac{x (840 x^{6} - 1540 x^{4} + 1022 x^{2} - 279)}{384 {(2 x^{2} - 1)}^{4}} + \frac{35}{128} \int \frac{d x}{2 x^{2} - 1} =

= \frac{x (840 x^{6} - 1540 x^{4} + 1022 x^{2} - 279)}{384 {(2 x^{2} - 1)}^{4}} + \frac{35 \sqrt{2}}{512} \ln \frac{\sqrt{2} x - 1}{\sqrt{2} x + 1} + C

\Rightarrow

\underset{1}{\int^{\infty}} \frac{d x}{{(2 x^{2} - 1)}^{5}} = - \frac{43}{384} + \frac{35 \sqrt{2}}{256} \ln (1 + \sqrt{2})

Commented by Bird last updated on 02/Oct/20

t h a n k y o u s i r

Answered by 1549442205PVT last updated on 03/Oct/20

\frac{2}{{2 x}^{2} - 1} = \frac{1}{\sqrt{2} x - 1} - \frac{1}{\sqrt{2} x + 1} . Hence,

Put \frac{1}{\sqrt{2} x - 1} = a, \frac{1}{\sqrt{2} x + 1} = b \Rightarrow 2 ab = a - b

\frac{1}{{({2 x}^{2} - 1)}^{5}} = \frac{1}{32} \times {(\frac{2}{{2 x}^{2} - 1})}^{5} = \frac{1}{32} {(a - b)}^{5}

We have {(a - b)}^{5}

= a^{5} - b^{5} - 5 ab (a^{3} - b^{3}) + 10 {(ab)}^{2} (a - b)

a^{5} - b^{5} - \frac{5}{2} (a - b) (a^{3} - b^{3}) + \frac{10}{4} {(a - b)}^{3}

= a^{5} - b^{5} - \frac{5}{2} (a^{4} + b^{4}) + \frac{5}{2} ab (a^{2} + b^{2}) + \frac{5}{2} [a^{3} - b^{3} - 3 ab (a - b)]

= a^{5} - b^{5} - \frac{5}{2} (a^{4} + b^{4}) + \frac{5}{4} (a - b) (a^{2} + b^{2}) + \frac{5}{2} [a^{3} - b^{3} - 3 ab (a - b)]]

= a^{5} - b^{5} - \frac{5}{2} (a^{4} + b^{4}) + \frac{5}{4} (a^{3} - b^{3}) - \frac{5}{4} ab (a - b) + \frac{5}{2} [a^{3} - b^{3} - 3 ab (a - b)]]

= a^{5} - b^{5} - \frac{5}{2} (a^{4} + b^{4}) + \frac{15}{4} (a^{3} - b^{3}) - \frac{35}{8} {(a - b)}^{2}

= a^{5} - b^{5} - \frac{5}{2} (a^{4} + b^{4}) + \frac{15}{4} (a^{3} - b^{3}) - \frac{35}{8} (a^{2} + b^{2}) + \frac{35}{8} (a - b)

I = \int_{1}^{\infty} [\frac{1}{32 {(\sqrt{2} x - 1)}^{5}} - \frac{1}{32 {(\sqrt{2} x + 1)}^{5}} - \frac{5}{64 {(\sqrt{2} x - 1)}^{4}}

- \frac{5}{64 {(\sqrt{2} x + 1)}^{4}} + \frac{15}{128 {(\sqrt{2} x - 1)}^{3}} - \frac{15}{128 {(\sqrt{2} x + 1)}^{3}}

- \frac{35}{256 {(\sqrt{2} x - 1)}^{2}} - \frac{35}{256 {(\sqrt{2} x + 1)}^{2}} + \frac{35}{256} (a - b)] dx

Put \sqrt{2} x - 1 = u, \sqrt{2} x + 1 = v \Rightarrow \sqrt{2} dx = du = dv . Hence,

I = - \frac{1}{128 \sqrt{2} {(\sqrt{2} x - 1)}^{4}} + \frac{1}{128 \sqrt{2} {(\sqrt{2} x + 1)}^{4}}

+ \frac{5}{192 \sqrt{2} {(\sqrt{2} x - 1)}^{3}} + \frac{5}{192 \sqrt{2} {(\sqrt{2} x + 1)}^{3}}

- \frac{15}{256 \sqrt{2} {(\sqrt{2} x - 1)}^{2}} + \frac{15}{256 \sqrt{2} {(\sqrt{2} x + 1)}^{2}}

{+ \frac{35}{256 \sqrt{2} (\sqrt{2} x - 1)} + \frac{35}{256 \sqrt{2} (\sqrt{2} x + 1)}]}_{1}^{\infty}

+ \frac{35}{256 \sqrt{2}} \ln ∣ \frac{\sqrt{2} x - 1}{\sqrt{2} x + 1} ∣

= [\frac{- (16 \sqrt{2} x^{3} + 8 \sqrt{2} x)}{128 \sqrt{2} {({2 x}^{2} - 1)}^{4}} + \frac{5 (4 \sqrt{2} x^{3} + 6 \sqrt{2} x)}{192 \sqrt{2} {({2 x}^{2} - 1)}^{3}}

{- \frac{15 \times (4 \sqrt{2} x)}{256 \sqrt{2} {({2 x}^{2} - 1)}^{2}} + \frac{35 \times 2 \sqrt{2} x}{256 \sqrt{2} ({2 x}^{2} - 1)} + \frac{35}{256 \sqrt{2}} \ln ∣ \frac{\sqrt{2} x - 1}{\sqrt{2} x + 1} ∣]}_{1}^{\infty}

= - (\frac{- 24 \sqrt{2}}{128 \sqrt{2}} + \frac{50 \sqrt{2}}{192 \sqrt{2}} - \frac{60 \sqrt{2}}{256 \sqrt{2}} + \frac{70 \sqrt{2}}{256 \sqrt{2}} + \frac{35}{256} \ln \frac{\sqrt{2} - 1}{\sqrt{2} + 1})

= - \frac{43}{384} - \frac{35 \sqrt{2}}{256} \ln (\sqrt{2} - 1) \approx 0.058434

Commented by Bird last updated on 02/Oct/20

t h a n k y o u s i r f o r t h i s h a r d w o r k

Commented by 1549442205PVT last updated on 03/Oct/20

Thank Sir . You are welcome .

Answered by Bird last updated on 03/Oct/20

I = \int_{1}^{\infty} \frac{d x}{{(2 x^{2} - 1)}^{5}} \Rightarrow

I = \int_{1}^{\infty} \frac{d x}{{(x \sqrt{2} - 1)}^{5} {(x \sqrt{2} + 1)}^{5}}

= \int_{1}^{\infty} \frac{d x}{{(\frac{x \sqrt{2} - 1}{x \sqrt{2} + 1})}^{5} {(x \sqrt{2} + 1)}^{10}}

w e d o t h e c h a n g e m e n t \frac{x \sqrt{2} - 1}{x \sqrt{2} + 1} = t

\Rightarrow x \sqrt{2} - 1 = t x \sqrt{2} + t \Rightarrow

(\sqrt{2} - t \sqrt{2}) x = t + 1 \Rightarrow x = \frac{t + 1}{\sqrt{2} (1 - t)}

\Rightarrow \frac{d x}{d t} = \frac{1}{\sqrt{2}} . \frac{1 - t - (t + 1) (- 1)}{{(1 - t)}^{2}}

= \frac{1}{\sqrt{2}} . \frac{2}{{(t - 1)}^{2}} a l s o x \sqrt{2} + 1 = \frac{t + 1}{1 - t} + 1

= \frac{t + 1 + 1 - t}{1 - t} = \frac{2}{1 - t} \Rightarrow

\Rightarrow I = \int_{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}^{1} \frac{1}{t^{5} {(\frac{2}{1 - t})}^{10}} \times \frac{\sqrt{2}}{{(t - 1)}^{2}} d t

= \frac{\sqrt{2}}{2^{10}} \int_{3 - 2 \sqrt{2}}^{1} \frac{{(t - 1)}^{8}}{t^{5}} d t

= \frac{\sqrt{2}}{2^{10}} \int_{3 - 2 \sqrt{2}}^{1} \frac{\sum_{k = 0}^{8} C_{8}^{k} t^{k} {(- 1)}^{8 - k}}{t^{5}} d t

= \frac{\sqrt{2}}{2^{10}} \sum_{k = 0}^{8} {(- 1)}^{k} C_{8}^{k} \int_{3 - 2 \sqrt{2}}^{1} t^{k - 5} d t

= \frac{\sqrt{2}}{2^{10}} \sum_{k = 0 a n d k \neq 4}^{8} {(- 1)}^{k} C_{8}^{k} {[\frac{1}{k - 4} t^{k - 4}]}_{3 - 2 \sqrt{2}}^{1}

+ \frac{\sqrt{2}}{2^{10}} C_{8}^{4} {[l n t]}_{3 - 2 \sqrt{2}}^{1}

I = \frac{\sqrt{2}}{2^{10}} \sum_{k = 0 a n d k \neq 4}^{8} \frac{{(- 1)}^{k} C_{8}^{k}}{k - 4} {1 - {(3 - 2 \sqrt{2})}^{k - 4}}

- \frac{\sqrt{2}}{2^{10}} C_{8}^{k} l n (3 - 2 \sqrt{2})

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