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

1 Answer
Jun 20, 2017

#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)#