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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#