Question

Question #6f0de

Answer 1

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

Explanation:

`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

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`