calculate ∫_2 ^∞ (dx/((x+1)^3 (x^2 +1)^4 ))


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Question Number 97624 by mathmax by abdo last updated on 08/Jun/20

$calculate \int_{2}^{\infty} \frac{dx}{{(x + 1)}^{3} {(x^{2} + 1)}^{4}}$
	Answered by MJS last updated on 09/Jun/20

	$\int \frac{d x}{{(x + 1)}^{3} {(x^{2} + 1)}^{4}} =$ $[Ostrogradski]$ $= - \frac{45 x^{7} - 6 x^{6} + 37 x^{5} - 16 x^{4} + - 101 x^{3} - 26 x^{2} - 109 x - 32}{192 {(x + 1)}^{2} {(x^{2} + 1)}^{3}} - \int \frac{15 x - 49}{64 (x + 1) (x^{2} + 1)} d x$ $- \int \frac{15 x - 49}{64 (x + 1) (x^{2} + 1)} d x = \frac{1}{2} \int \frac{d x}{x + 1} - \int \frac{32 x - 17}{64 (x^{2} + 1)} d x =$ $= \frac{1}{2} \ln (x + 1) - \frac{1}{4} \ln (x^{2} + 1) + \frac{17}{64} \arctan x =$ $= \frac{1}{4} \ln \frac{{(x + 1)}^{2}}{x^{2} + 1} + \frac{17}{64} \arctan x + C$ $\underset{2}{\int^{+ \infty}} \frac{d x}{{(x + 1)}^{3} {(x^{2} + 1)}^{4}} = \frac{857}{36000} + \frac{1}{4} \ln \frac{5}{9} + \frac{17}{64} \arctan \frac{1}{2}$
	Commented by I want to learn more last updated on 09/Jun/20

	$I appreciate sir .$
	Commented by MJS last updated on 09/Jun/20

	$I explained {Ostrogradski}^{'} s method a few$ $weeks ago, cannot find the post now . there$ $are explanations on the web$
	Commented by I want to learn more last updated on 09/Jun/20

	$Ohh . I wish you could find it sir, have missed$
	Commented by MJS last updated on 09/Jun/20

	$q u e s t i o n 94354$
	Answered by mathmax by abdo last updated on 09/Jun/20

	$complex method I = \int_{2}^{\infty} \frac{dx}{{(x + 1)}^{3} {(x^{2} + 1)}^{4}} \Rightarrow$ $I = \int_{2}^{\infty} \frac{dx}{{(x + 1)}^{3} {(x - i)}^{4} {(x + i)}^{4}} = \int_{2}^{\infty} \frac{dx}{{(\frac{x + 1}{x - i})}^{3} {(x - i)}^{7} {(x + i)}^{4}}$ $changement \frac{x + 1}{x - i} = t give x + 1 = tx - it \Rightarrow (1 - t) x = - 1 - it \Rightarrow x = \frac{- it - 1}{1 - t}$ $= \frac{it + 1}{t - 1} \Rightarrow \frac{dx}{dt} = \frac{i (t - 1) - it - 1}{{(t - 1)}^{2}} = \frac{- i - 1}{{(t - 1)}^{2}}$ $x - i = \frac{it + 1}{t - 1} - i = \frac{it + 1 - it + i}{t - 1} = \frac{1 + i}{t - 1}$ $x + i = \frac{it + 1}{t - 1} + i = \frac{it + 1 + it - i}{t - 1} = \frac{2 it + 1 - i}{t - 1} \Rightarrow$ $I = \int_{\frac{3}{2 - i}}^{1} \frac{(- 1 - i)}{{(t - 1)}^{2} t^{3} {(\frac{1 + i}{t - 1})}^{7} {(\frac{2 it + 1 - i}{t - 1})}^{4}} dt$ $= - \frac{1}{{(1 + i)}^{6}} \int_{\frac{3}{2 - i}}^{1} \frac{{(t - 1)}^{11}}{{(t - 1)}^{2} t^{3} {(2 it + 1 - i)}^{4}} dt$ $= - \frac{1}{{(1 + i)}^{6}} \int_{\frac{3}{2 i}}^{1} \frac{{(t - 1)}^{9}}{t^{3} {(2 it + 1 - i)}^{4}} dt but 2 it + 1 - i = 2 i (t + \frac{1 - i}{2})$ $\Rightarrow I = - \frac{1}{{(1 + i)}^{6} {(2 i)}^{4}} \int_{\frac{3}{2 i}}^{1} \frac{{(t - 1)}^{9}}{t^{3} {(t + \frac{1 - i}{2})}^{4}} dt$ $we have \int_{- \frac{3 i}{2}}^{1} \frac{{(t - 1)}^{9}}{t^{3} {(t + \frac{1 - i}{2})}^{4}} dt = \int_{- \frac{3 i}{2}}^{1} \frac{\sum_{k = 0}^{9} C_{9}^{k} t^{k} {(- 1)}^{9 - k}}{t^{3} {(t + \frac{1 - i}{2})}^{4}} dt$ $= - \sum_{k = 0}^{9} C_{9}^{k} {(- 1)}^{k} t^{k - 3} \frac{dt}{{(t + \frac{1 - i}{2})}^{4}} let$ $φ_{k} (t) = \frac{t^{k - 3}}{{(t + \frac{1 - i}{2})}^{4}} \Rightarrow φ_{k} (t) = \frac{a_{1}}{t + \frac{1 - i}{2}} + \frac{a_{2}}{{(t + \frac{1 - i}{2})}^{2}} + \frac{a_{3}}{{(t + \frac{1 - i}{2})}^{3}} + \frac{a_{3}}{{(t + \frac{1 - i}{2})}^{4}}$ $rest to calculate a_{i} . . . be continued . . .$
	Commented by I want to learn more last updated on 09/Jun/20

	$Thanks sir . Waiting sir$
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