Question #6f0de

2 Answers
Nov 22, 2017

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

Nov 22, 2017

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