Question

How do you determine if the improper integral converges or diverges `int 3/(7x^2 +4)` from negative infinity to infinity?

Answer 1

`int_(-oo)^(+oo) 3/(7x^2+4)dx = (3pi)/(2sqrt7)`

Explanation:

Given:

`f(x) = 3/(7x^2+4)`

Evaluate first:

`lim_(t->oo) int_0^t 3/(7x^2+4) dx = lim_(t->oo) 3/4 int_0^t dx/(7/4x^2+1)`

`lim_(t->oo) int_0^t 3/(7x^2+4) dx = lim_(t->oo) 3/(2sqrt7) int_0^t (d(sqrt7/2x))/( (sqrt7/2x)^2+1)`

`lim_(t->oo) int_0^t 3/(7x^2+4) dx = lim_(t->oo) 3/(2sqrt7) [arctan(sqrt7/2x)]_0^t`

`lim_(t->oo) int_0^t 3/(7x^2+4) dx = lim_(t->oo) 3/(2sqrt7) arctan(sqrt7/2t)`

`lim_(t->oo) int_0^t 3/(7x^2+4) dx = (3pi)/(4sqrt7)`

Then the integral:

`int_0^oo 3/(7x^2+4)dx = (3pi)/(4sqrt7)`

is convergent.

Evaluate then:

`int_(-t)^0 3/(7x^2+4)dx`

substitute now: `u=-x`:

`int_(-t)^0 3/(7x^2+4)dx = - int_t^0 3/(7(-u)^2+4)du = int_0^t 3/(7u^2+4)du`

so that also:

`int_(-oo)^0 3/(7x^2+4)dx = (3pi)/(4sqrt7)`

is convergent.

Finally then:

`int_(-oo)^(+oo) 3/(7x^2+4)dx = (3pi)/(2sqrt7)`