Question

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

Answer 1

Refer to explanation

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.