How do you show whether the improper integral #int (x^2)/(9+x^6) dx# converges or diverges from negative infinity to infinity?

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Answer 1 · 2015-09-27T22:15:19

Refer to explanation

Let's calculate the indefinite integral hence we have

#int x^2/(x^6+9)dx=1/3int ((x^3)')/(9+(x^3)^2)dx=1/9arctan(x^3/9)+c#

Hence the definite is

#int_(-oo)^(+oo) x^2/(x^6+9)dx=lim_(x->+oo)F(x)-lim_(x->-oo)F(x)#

where #F(x)=1/9arctan(x^3/9)+c#

The arctan function is the inverse function of
#tan:(−π/2,π/2)→R#

and since this function is monotonically increasing then
#lim_(x->pi/2)tanx=+∞⟺lim_(x->+oo)arctanx=π/2#

So we have that

#int_(-oo)^(+oo) x^2/(x^6+9)dx=lim_(x->+oo)F(x)-lim_(x->-oo)F(x)=(pi/18)-(-pi/18)=pi/9#

Hence the integral converges.