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

Answer 1

# - 1/27pi #

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 #
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Answer 2

To solve the integral ( \int_{-\infty}^{+\infty} \frac{x}{(x^2 + 4x + 13)^2} , dx ), we can use the method of completing the square and partial fractions. After completing the square in the denominator, we can express the integrand as a sum of partial fractions. This allows us to integrate each term separately. The solution involves decomposition of the fraction into partial fractions and then integrating each term. The result is a combination of logarithmic and inverse trigonometric functions.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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