calculate-2-dx-x-3-x-2-1-2-

Question Number 99239 by abdomathmax last updated on 19/Jun/20

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

Answered by MJS last updated on 19/Jun/20

Ostrogradski gives

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

= - \frac{2 x^{2} - 1}{2 x^{2} (x^{2} - 1)} + \ln ∣ \frac{x^{2}}{x^{2} - 1} ∣ + C

\Rightarrow

\underset{2}{\int^{\infty}} \frac{d x}{x^{3} (x^{2} - 1)} = \frac{7}{24} + \ln \frac{3}{4}

Commented by Ar Brandon last updated on 20/Jun/20

Hello Mr MJS,�� greetings to you Sir.

Commented by MJS last updated on 20/Jun/20

also back to you!

Answered by mathmax by abdo last updated on 21/Jun/20

I = \int_{2}^{+ \infty} \frac{dx}{x^{3} {(x^{2} - 1)}^{2}} \Rightarrow I = \int_{2}^{\infty} \frac{dx}{{(\frac{x}{x - 1})}^{3} {(x - 1)}^{5} {(x + 1)}^{2}}

we do the changement \frac{x}{x - 1} = t \Rightarrow x = tx - t \Rightarrow (1 - t) x = - t \Rightarrow x = \frac{t}{t - 1}

\Rightarrow \frac{dx}{dt} = {(1 + \frac{1}{t - 1})}^{^{'}} = - \frac{1}{{(t - 1)}^{2}} and x - 1 = \frac{t}{t - 1} - 1 = \frac{t - t + 1}{t - 1} = \frac{1}{t - 1}

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

I = - \int_{1}^{2} \frac{- dt}{{(t - 1)}^{2} t^{3} {(\frac{1}{t - 1})}^{5} {(\frac{2 t - 1}{t - 1})}^{2}} = \int_{1}^{2} \frac{{(t - 1)}^{7}}{{(t - 1)}^{2} t^{3} {(2 t - 1)}^{2}} dt

= \int_{1}^{2} \frac{{(t - 1)}^{5}}{t^{3} {(2 t - 1)}^{2}} dt = \int_{1}^{2} \frac{\sum_{k = 0}^{5} C_{5}^{k} t^{k} {(- 1)}^{5 - k}}{t^{3} {(2 t - 1)}^{2}} dt

= - \sum_{k = 0}^{5} C_{5}^{k} {(- 1)}^{k} \int_{1}^{2} \frac{t^{k}}{t^{3} {(2 t - 1)}^{2}} dt after we decompose F (t) = \frac{t^{k}}{t^{3} {(2 t - 1)}^{2}}

\dots be continued \dots