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Question Number 33334 by prof Abdo imad last updated on 14/Apr/18
decompose F(x)= (x^2 /(x^4 −1)) imside R(x) 2) find the value of ∫_2 ^(+∞) (x^2 /(x^4 −1)) .
$${decompose}\:{F}\left({x}\right)=\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:−\mathrm{1}}\:{imside}\:{R}\left({x}\right)\: \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\:\:\int_{\mathrm{2}} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:−\mathrm{1}}\:\:. \\ $$
Commented by prof Abdo imad last updated on 15/Apr/18
2) find the value of ∫_2 ^(+∞) (x^2 /(x^4 −1)) dx.
$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{2}} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:−\mathrm{1}}\:\:{dx}. \\ $$
Commented by prof Abdo imad last updated on 18/Apr/18
we have F(x)= (x^2 /((x^2 −1)(x^2 +1))) =((x^2 −1 +1)/((x^2 −1)(x^2 +1))) = (1/(x^2 +1)) + (1/((x^2 −1)(x^2 +1))) let decompose g(x)= (1/((x^2 −1)(x^2 +1))) g(x) = (a/(x−1)) +(b/(x+1)) + ((cx+d)/(x^2 +1)) g(−x)=g(x)⇔ ((−a)/(x+1)) +((−b)/(x−1)) +((−cx +d)/(x^2 +1)) = g(x) ⇒ b=−a and c=0 ⇒ g(x)= (a/(x−1)) −(a/(x+1)) + (d/(x^2 +1)) g(0) = −1 = −2a +d ⇒ 2a −d =1 g(2) = (1/(3 .5)) =(1/(15)) = a −(a/3) +(d/5) = ((2a)/3) +(d/5) ⇒ 1 = 10a +3d ⇒ 10a +3(2a−1) =1 ⇒ 16a =4 ⇒ a = (1/4) and d =(1/2) −1=−(1/2) g(x) = (1/(4(x−1))) −(1/(4(x+1))) − (1/(2(x^2 +1))) ⇒ F(x)= (1/(x^2 +1)) +(1/(4(x−1))) −(1/(4(x+1))) −(1/(2(x^2 +1))) F(x)= (1/(2(x^2 +1))) + (1/(4(x−1))) −(1/(4(x+1)))
$${we}\:{have}\:{F}\left({x}\right)=\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:−\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} \:−\mathrm{1}\:+\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${let}\:{decompose}\:{g}\left({x}\right)=\:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${g}\left({x}\right)\:=\:\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{{x}+\mathrm{1}}\:\:+\:\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${g}\left(−{x}\right)={g}\left({x}\right)\Leftrightarrow\:\frac{−{a}}{{x}+\mathrm{1}}\:\:\:+\frac{−{b}}{{x}−\mathrm{1}}\:\:+\frac{−{cx}\:+{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\:=\:{g}\left({x}\right)\:\Rightarrow\:{b}=−{a}\:\:{and}\:{c}=\mathrm{0}\:\:\Rightarrow \\ $$$${g}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:−\frac{{a}}{{x}+\mathrm{1}}\:\:+\:\frac{{d}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${g}\left(\mathrm{0}\right)\:=\:−\mathrm{1}\:=\:−\mathrm{2}{a}\:+{d}\:\:\Rightarrow\:\mathrm{2}{a}\:−{d}\:=\mathrm{1} \\ $$$${g}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}\:.\mathrm{5}}\:=\frac{\mathrm{1}}{\mathrm{15}}\:=\:{a}\:−\frac{{a}}{\mathrm{3}}\:+\frac{{d}}{\mathrm{5}}\:=\:\frac{\mathrm{2}{a}}{\mathrm{3}}\:+\frac{{d}}{\mathrm{5}}\:\Rightarrow \\ $$$$\mathrm{1}\:=\:\mathrm{10}{a}\:+\mathrm{3}{d}\:\:\Rightarrow\:\mathrm{10}{a}\:+\mathrm{3}\left(\mathrm{2}{a}−\mathrm{1}\right)\:=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{16}{a}\:=\mathrm{4}\:\Rightarrow\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:{and}\:{d}\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left({x}\right)\:=\:\:\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)}\:−\:\frac{\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:+\:\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)} \\ $$
Commented by math khazana by abdo last updated on 19/Apr/18
∫_2 ^(+∞) (x^2 /(x^4 −1))dx = (1/2) ∫_2 ^(+∞) (dx/(1+x^2 )) +(1/4) ∫_2 ^(+∞) ((1/(x−1)) −(1/(x+1)))dx = (1/2)[ arctanx^ ]_2 ^(+∞) +(1/4) [ ln∣((x−1)/(x+1))∣]_2 ^(+∞) =(1/2)( (π/2) −arctan2) +(1/4)(−ln((1/3))) =(π/4) −(1/2) arctan(2) +(1/4)ln(3) .
$$\int_{\mathrm{2}} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} −\mathrm{1}}{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{2}} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{2}} ^{+\infty} \:\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\:{arctanx}^{} \right]_{\mathrm{2}} ^{+\infty} \:\:\:+\frac{\mathrm{1}}{\mathrm{4}}\:\left[\:{ln}\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\mid\right]_{\mathrm{2}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\frac{\pi}{\mathrm{2}}\:−{arctan}\mathrm{2}\right)\:\:\:+\frac{\mathrm{1}}{\mathrm{4}}\left(−{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\mathrm{2}\right)\:\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{3}\right)\:\:. \\ $$

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