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Answer 1 · 2017-11-22T19:02:34

$int_-oo^oo x^2/(x^6+9)*dx=pi/9$

$int_-oo^oo x^2/(x^6+9)*dx$

= $1/3*int_-oo^oo (3x^2*dx)/[(x^3)^2+9]$

After using $y=x^3$ an $dy=3x^2*dx$ transformation, this integral became

$1/3*int_-oo^oo dy/(y^2+9)$

= $2/3*int_0^oo dy/(y^2+9)$

= $2/9*[arctan(y/3)]_0^oo$

= $2/9*[arctan(oo)-arctan(0)]$

= $2/9*(pi/2-0)$

= $pi/9$

Answer 2 · 2017-11-22T23:56:50

2nd way: I didn't use symmetry

$int_-oo^oo x^2/(x^6+9)*dx$

= $1/3*int_-oo^oo (3x^2*dx)/[(x^3)^2+9]$

After using $y=x^3$ an $dy=3x^2*dx$ transformation, this integral became

$1/3*int_-oo^oo dy/(y^2+9)$

Now I divided range of integral. First of it from $-oo$ to $a$ and second of it from $a$ to $oo$ .

$1/3*int_-oo^a dy/(y^2+9)+1/3*int_a^oo dy/(y^2+9)$

= $1/9*[arctan(y/3)]_-oo^a+1/9*[arctan(y/3)]_a^oo$

= $1/9*[arctan(a/3)-arctan(-oo)]+1/9*[arctan(oo)-arctan(a/3)]$

= $1/9*[arctan(oo)-arctan(-oo)]$

= $1/9*[pi/2-(-pi/2)]$

= $pi/9$