Question

How to solve this integral `int_(-oo)^(+oo)(x)/((x^2+4x+13)^2)dx`?

Answer 1

`- 1/27pi`

Explanation:

Let:

`I = int_(-oo)^(oo) \ x/(x^2+4x+13)^2 \ dx`

We attempt to manipulate the numerator so that it contains the derivative of the quadratic in the denominator, ie `2x+4`

`I = int_(-oo)^(oo) \ (1/2(2x+4)-2)/(x^2+4x+13)^2 \ dx`
`\ \ = int_(-oo)^(oo) \ (1/2(2x+4))/(x^2+4x+13)^2 - (2)/(x^2+4x+13)^2\ dx`

Consider the first integral (we will deal with the improper integral)

`I_1 = int \ (2x+4)/(x^2+4x+13)^2 \ dx`

We can perform a simple substitution:
Let `u=x^2+4x+13 => (du)/dx=2x+3`

`:. I_1 = int \ 1/u^2 \ du`
`" " = -1/u`
`" " = -1/(x^2+4x+13)`

Now, consider the second integral (again we will deal with the improper integral)

`I_2 = int \ 1/(x^2+4x+13)^2\ dx`

We can complete the square on the denominator to get:

`I_2 = int \ 1/((x+2)^2-2^2+13)^2\ dx`
`\ \ \ = int \ 1/((x+2)^2+9)^2\ dx`
`\ \ \ = int \ 1/((x+2)^2+3^2)^2\ dx`

Let `u=x+2 => (du)/dx=1`

`:. I_2 = int \ 1/(u^2+3^2)^2 \ du`

I'll just quote this result here, as the integration is a bit tedious:

`int 1/(u^2+a^2)^2 \ du = u/(2a^2(u^2+a^2)) + 1/(2a^3)tan^(-1) (u/a)`

`:. I_2 = u/(18(u^2+9)) + 1/54tan^(-1) (u/3)`
`" " = (x+2)/(18((x+2)^2+9)) + 1/54tan^(-1) ((x+2)/3)`

Combining our results we get:

`I = -1/2 * [1/(x^2+4x+13) ]_(-oo)^(oo)`
`" "- 2*[(x+2)/(18((x+2)^2+9)) + 1/54tan^(-1) ((x+2)/3)]_(-oo)^(oo)`

`\ \ = 0 - 2 * [0 + 1/54tan^(-1) ((x+2)/3)]_(-oo)^(oo)`

`\ \ = - 1/27[tan^(-1) ((x+2)/3)]_(-oo)^(oo)`

`\ \ = - 1/27(pi/2-(-pi/2)`
`\ \ = - 1/27pi`