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Evaluate \( \int \frac{-4 x-4}{x^{2}+1} d x= \) (Don't forget to include the constant of antidifferentiation \( c \) )

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Answer:

The evaluated integral is ∫(-4x - 4)/(x^2 + 1) dx = -2ln|x + i| - 2ln|x - i| + C

Explanation:

To evaluate the integral ∫(-4x - 4)/(x^2 + 1) dx, we can use the method of partial fractions.

First, let's rewrite the integrand as a sum of two fractions:

(-4x - 4)/(x^2 + 1) = (-4x)/(x^2 + 1) - 4/(x^2 + 1)

Now, let's find the partial fraction decomposition. We can express the integrand as:

(-4x)/(x^2 + 1) - 4/(x^2 + 1) = A/(x + i) + B/(x - i)

where A and B are constants to be determined. Here, i represents the imaginary unit (√(-1)).

To find A and B, we can multiply both sides of the equation by x^2 + 1 and then equate the numerators:

-4x - 4 = A(x - i) + B(x + i)

Expanding and simplifying:

-4x - 4 = (A + B)x + (B - A)i

Equating the real and imaginary parts separately:

-4x = (A + B)x (equation 1)

-4 = (B - A)i (equation 2)

From equation 1, we have A + B = -4.

From equation 2, we have B - A = 0, which implies B = A.

Substituting B = A into A + B = -4, we get 2A = -4, which gives A = -2. Consequently, B = -2 as well.

Now, we can rewrite the integral using the partial fraction decomposition:

∫(-4x - 4)/(x^2 + 1) dx = ∫(-2/(x + i) - 2/(x - i)) dx

Integrating each term separately:

= -2∫1/(x + i) dx - 2∫1/(x - i) dx

To integrate 1/(x + i), we can use the natural logarithm:

= -2ln|x + i| - 2ln|x - i| + C

where C is the constant of integration.

Therefore,

∫(-4x - 4)/(x^2 + 1) dx = -2ln|x + i| - 2ln|x - i| + C is the integral.

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