How to solve this integral int_(-oo)^(+oo)(x)/((x^2+4x+13)^2)dx?
1 Answer
- 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
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
:. 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
:. 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