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Question Number 33334 by prof Abdo imad last updated on 14/Apr/18

$$decomposeF\left(x\right)=\frac{x^2}{x^4-1}imsideR\left(x\right)$$$$2)findthevalueof{\int }_2^{+\mathrm{\infty }}\frac{x^2}{x^4-1}.$$

Commented by prof Abdo imad last updated on 15/Apr/18

$$2)findthevalueof{\int }_2^{+\mathrm{\infty }}\frac{x^2}{x^4-1}dx.$$

Commented by prof Abdo imad last updated on 18/Apr/18

$$wehaveF\left(x\right)=\frac{x^2}{(x^2-1)(x^2+1)}$$$$=\frac{x^2-1+1}{(x^2-1)(x^2+1)}=\frac{1}{x^2+1}+\frac{1}{(x^2-1)(x^2+1)}$$$$letdecomposeg\left(x\right)=\frac{1}{(x^2-1)(x^2+1)}$$$$g\left(x\right)=\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}$$$$g(-x)=g\left(x\right)\iff \frac{-a}{x+1}+\frac{-b}{x-1}+\frac{-cx+d}{x^2+1}$$$$=g\left(x\right)\Rightarrow b=-aandc=0\Rightarrow$$$$g\left(x\right)=\frac{a}{x-1}-\frac{a}{x+1}+\frac{d}{x^2+1}$$$$g\left(0\right)=-1=-2a+d\Rightarrow 2a-d=1$$$$g\left(2\right)=\frac{1}{3.5}=\frac{1}{15}=a-\frac{a}{3}+\frac{d}{5}=\frac{2a}{3}+\frac{d}{5}\Rightarrow$$$$1=10a+3d\Rightarrow 10a+3(2a-1)=1\Rightarrow$$$$16a=4\Rightarrow a=\frac{1}{4}andd=\frac{1}{2}-1=-\frac{1}{2}$$$$g\left(x\right)=\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2(x^2+1)}\Rightarrow$$$$F\left(x\right)=\frac{1}{x^2+1}+\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2(x^2+1)}$$$$F\left(x\right)=\frac{1}{2(x^2+1)}+\frac{1}{4(x-1)}-\frac{1}{4(x+1)}$$

Commented by math khazana by abdo last updated on 19/Apr/18

$${\int }_2^{+\mathrm{\infty }}\frac{x^2}{x^4-1}dx=\frac{1}{2}{\int }_2^{+\mathrm{\infty }}\frac{dx}{1+x^2}+\frac{1}{4}{\int }_2^{+\mathrm{\infty }}(\frac{1}{x-1}-\frac{1}{x+1})dx$$$$=\frac{1}{2}{\left[{arctanx}^{}\right]}_2^{+\mathrm{\infty }}+\frac{1}{4}{[ln\mid \frac{x-1}{x+1}\mid ]}_2^{+\mathrm{\infty }}$$$$=\frac{1}{2}(\frac{\pi }{2}-arctan2)+\frac{1}{4}(-ln\left(\frac{1}{3}\right))$$$$=\frac{\pi }{4}-\frac{1}{2}arctan\left(2\right)+\frac{1}{4}ln\left(3\right).$$

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