please help ∫(1/(x^2 (x^2 −2)^2 ))dx=?


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Question Number 129938 by stelor last updated on 21/Jan/21

$please help \int \frac{1}{x^{2} {(x^{2} - 2)}^{2}} d x = ?$
	Answered by Ar Brandon last updated on 21/Jan/21
	$I=∫(dx/(x^2 (2−x^2 )^2 ))=(1/2)∫(((2−x^2 )+x^2 )/(x^2 (2−x^2 )^2 ))dx=(1/2)∫{(1/(x^2 (2−x^2 )))+(1/((2−x^2 )^2 ))}dx =(1/4)∫(((2−x^2 )+x^2 )/(x^2 (2−x^2 )))dx+(1/2)∫(dx/((2−x^2 )^2 )) =(1/4)∫{(1/x^2 )+(1/(2−x^2 ))}+(1/2)∫(dx/((2−x^2 )^2 )) f(a)=∫(dx/(a^2 −x^2 ))=(1/a)arctanh((x/a)) f ′(a)=−∫((2a)/((a^2 −x^2 )^2 ))dx=(1/a)∙(−(x/a^2 )∙(1/(1−(x^2 /a^2 ))))−(1/a^2 )arctanh((x/a)) ⇒∫(dx/((2−x^2 )^2 ))=−(1/(2(√2)))f ′((√2)) I=(1/4){−(1/x)+(1/( (√2)))tanh^(−1) ((x/( (√2))))}−(1/(4(√2))){(1/( (√2)))∙(x/(x^2 −2))−(1/2)tanh^(−1) ((x/( (√2))))} =(3/(8(√2)))tanh^(−1) ((x/( (√2))))−(1/8)∙(x/(x^2 −2))−(1/(4x))+C$
	$I = \int \frac{dx}{x^{2} {(2 - x^{2})}^{2}} = \frac{1}{2} \int \frac{(2 - x^{2}) + x^{2}}{x^{2} {(2 - x^{2})}^{2}} dx = \frac{1}{2} \int {\frac{1}{x^{2} (2 - x^{2})} + \frac{1}{{(2 - x^{2})}^{2}}} dx$ $= \frac{1}{4} \int \frac{(2 - x^{2}) + x^{2}}{x^{2} (2 - x^{2})} dx + \frac{1}{2} \int \frac{dx}{{(2 - x^{2})}^{2}}$ $= \frac{1}{4} \int {\frac{1}{x^{2}} + \frac{1}{2 - x^{2}}} + \frac{1}{2} \int \frac{dx}{{(2 - x^{2})}^{2}}$ $f (a) = \int \frac{dx}{a^{2} - x^{2}} = \frac{1}{a} arctanh (\frac{x}{a})$ $f^{'} (a) = - \int \frac{2 a}{{(a^{2} - x^{2})}^{2}} dx = \frac{1}{a} \cdot (- \frac{x}{a^{2}} \cdot \frac{1}{1 - \frac{x^{2}}{a^{2}}}) - \frac{1}{a^{2}} arctanh (\frac{x}{a})$ $\Rightarrow \int \frac{dx}{{(2 - x^{2})}^{2}} = - \frac{1}{2 \sqrt{2}} f^{'} (\sqrt{2})$ $I = \frac{1}{4} {- \frac{1}{x} + \frac{1}{\sqrt{2}} \tanh^{- 1} (\frac{x}{\sqrt{2}})} - \frac{1}{4 \sqrt{2}} {\frac{1}{\sqrt{2}} \cdot \frac{x}{x^{2} - 2} - \frac{1}{2} \tanh^{- 1} (\frac{x}{\sqrt{2}})}$ $= \frac{3}{8 \sqrt{2}} \tanh^{- 1} (\frac{x}{\sqrt{2}}) - \frac{1}{8} \cdot \frac{x}{x^{2} - 2} - \frac{1}{4 x} + C$
	Commented by stelor last updated on 21/Jan/21

	$merci - -$
	Answered by Ar Brandon last updated on 21/Jan/21
	$I=∫(dx/(x^2 (x^2 −2)^2 )) , x=(√2)sinθ =∫(((√2)cosθdθ)/(2sin^2 θ∙4(sin^2 θ−1)^2 ))=((√2)/8)∫(dθ/(sin^2 θcos^3 θ)) =((√2)/( 8))∫{(1/(sin^2 θcosθ))+(1/(cos^3 θ))}dθ=((√2)/8)∫{((cosθ)/(sin^2 θ))+(1/(cosθ))+(1/(cos^3 θ))}dθ =((√2)/8){−(1/(sinθ))+ln∣tan((θ/2)+(π/4))∣}+((√2)/( 8))∫(dθ/(cos^3 θ)) ∫(dθ/(cos^3 θ))=∫((cosθ)/(cos^4 θ))dθ=∫((cosθ)/((1−sin^2 θ)^2 ))dθ =∫(du/((1−u^2 )^2 ))=∫(du/(u^4 −2u^2 +1))=(1/2)∫{(((u^2 +1)−(u^2 −1))/(u^4 −2u^2 +1))}du =(1/2)∫{((1+(1/u^2 ))/((u−(1/u))^2 ))−((1−(1/u^2 ))/((u+(1/u))^2 −4))}du =(1/2){∫(dp/p^2 )−∫(dq/(q^2 −4))}=(1/2){−(1/p)+(1/4)ln∣((q+2)/(q−2))∣}+C$
	$I = \int \frac{dx}{x^{2} {(x^{2} - 2)}^{2}}, x = \sqrt{2} \sin θ$ $= \int \frac{\sqrt{2} \cos θ d θ}{{2 \sin}^{2} θ \cdot 4 {(\sin^{2} θ - 1)}^{2}} = \frac{\sqrt{2}}{8} \int \frac{d θ}{\sin^{2} θ \cos^{3} θ}$ $= \frac{\sqrt{2}}{8} \int {\frac{1}{\sin^{2} θ \cos θ} + \frac{1}{\cos^{3} θ}} d θ = \frac{\sqrt{2}}{8} \int {\frac{\cos θ}{\sin^{2} θ} + \frac{1}{\cos θ} + \frac{1}{\cos^{3} θ}} d θ$ $= \frac{\sqrt{2}}{8} {- \frac{1}{\sin θ} + \ln ∣ \tan (\frac{θ}{2} + \frac{π}{4}) ∣} + \frac{\sqrt{2}}{8} \int \frac{d θ}{\cos^{3} θ}$ $\int \frac{d θ}{\cos^{3} θ} = \int \frac{\cos θ}{\cos^{4} θ} d θ = \int \frac{\cos θ}{{(1 - \sin^{2} θ)}^{2}} d θ$ $= \int \frac{du}{{(1 - u^{2})}^{2}} = \int \frac{du}{u^{4} - {2 u}^{2} + 1} = \frac{1}{2} \int {\frac{(u^{2} + 1) - (u^{2} - 1)}{u^{4} - {2 u}^{2} + 1}} du$ $= \frac{1}{2} \int {\frac{1 + \frac{1}{u^{2}}}{{(u - \frac{1}{u})}^{2}} - \frac{1 - \frac{1}{u^{2}}}{{(u + \frac{1}{u})}^{2} - 4}} du$ $= \frac{1}{2} {\int \frac{dp}{p^{2}} - \int \frac{dq}{q^{2} - 4}} = \frac{1}{2} {- \frac{1}{p} + \frac{1}{4} \ln ∣ \frac{q + 2}{q - 2} ∣} + C$
	Commented by liberty last updated on 21/Jan/21

	$x = \sqrt{2} \sin θ \Rightarrow dx = \sqrt{2} \cos θ d θ$
	Commented by Ar Brandon last updated on 21/Jan/21
	Yeah, you're right. Thanks for remark.
	Answered by MJS_new last updated on 21/Jan/21

	$\int \frac{d x}{x^{2} {(x^{2} - 2)}^{2}} =$ $[{Ostrogradski}^{'} s Method]$ $= - \frac{3 x^{2} - 4}{8 x (x^{2} - 2)} - \frac{3}{8} \int \frac{d x}{x^{2} - 2} =$ $= - \frac{3 x^{2} - 4}{8 x (x^{2} - 2)} - \frac{3 \sqrt{2}}{32} \int \frac{1}{x + \sqrt{2}} - \frac{1}{x - \sqrt{2}} d x =$ $= - \frac{3 x^{2} - 4}{8 x (x^{2} - 2)} - \frac{3 \sqrt{2}}{32} (\ln ∣ x + \sqrt{2} ∣ - \ln ∣ x - \sqrt{2} ∣) =$ $= - \frac{3 x^{2} - 4}{8 x (x^{2} - 2)} - \frac{3 \sqrt{2}}{32} \ln ∣ \frac{x + \sqrt{2}}{x - \sqrt{2}} ∣ + C$
	Answered by mathmax by abdo last updated on 21/Jan/21

	$I = \int \frac{dx}{x^{2} {(x^{2} - 2)}^{2}} \Rightarrow I = \int \frac{dx}{x^{2} {(x - \sqrt{2})}^{2} {(x + \sqrt{2})}^{2}}$ $= \int \frac{dx}{x^{2} {(\frac{x - \sqrt{2}}{x + \sqrt{2}})}^{2} {(x + \sqrt{2})}^{4}} we do the changement \frac{x - \sqrt{2}}{x + \sqrt{2}} = t \Rightarrow$ $x - \sqrt{2} = tx + \sqrt{2} t \Rightarrow (1 - t) x = \sqrt{2} + \sqrt{2} t \Rightarrow x = \frac{\sqrt{2} t + \sqrt{2}}{1 - t} \Rightarrow$ $\frac{dx}{dt} = \frac{\sqrt{2} (1 - t) + \sqrt{2} t + \sqrt{2}}{{(1 - t)}^{2}} = \frac{2 \sqrt{2}}{{(1 - t)}^{2}} and x + \sqrt{2} = \frac{\sqrt{2} t + \sqrt{2}}{1 - t} + \sqrt{2}$ $= \frac{\sqrt{2} t + \sqrt{2} + \sqrt{2} - \sqrt{2} t}{1 - t} = \frac{2 \sqrt{2}}{1 - t} \Rightarrow$ $I = \int \frac{2 \sqrt{2}}{{(1 - t)}^{2} {(\frac{\sqrt{2} t + \sqrt{2}}{1 - t})}^{2} {(\frac{2 \sqrt{2}}{1 - t})}^{4}} dt$ $= \frac{1}{{(2 \sqrt{2})}^{2} \times 2} \int \frac{{(1 - t)}^{6}}{{(1 - t)}^{2} {(t + 1)}^{2}} dt$ $= \frac{1}{16} \int \frac{{(t - 1)}^{4}}{{(t + 1)}^{2}} dt =_{t + 1 = z} \frac{1}{16} \int \frac{{(z - 2)}^{4}}{z^{2}} dz$ $= \frac{1}{16} \int \frac{\sum_{k = 0}^{4} C_{4}^{k} z^{k} {(- 2)}^{4 - k}}{z^{2}} dz$ $= \frac{1}{16} \int {(- 2)}^{4} \sum_{k = 0}^{4} {(- 2)}^{k} C_{4}^{k} z^{k - 2} dz$ $= \sum_{k = 0 and k \neq 1}^{4} {(- 2)}^{k} C_{4}^{k} \frac{1}{k - 1} z^{k - 1} - 2 \ln ∣ z ∣ + c$ $= \sum_{k = 0}^{4} \frac{{(- 2)}^{k} C_{4}^{k}}{k - 1} {(t + 1)}^{k - 1} - 2 \ln ∣ t + 1 ∣ + C$ $= \sum_{k = 0}^{4} \frac{{(- 2)}^{k} C_{4}^{k}}{k - 1} {(\frac{x - \sqrt{2}}{x + \sqrt{2}} + 1)}^{k - 1} - 2 \ln ∣ \frac{x - \sqrt{2}}{3 + \sqrt{2}} + 1 ∣ + C$
	Commented by mathmax by abdo last updated on 21/Jan/21

	$k \neq 1$
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