52.9k views
1 vote
Evaluate \( \int \frac{-4 x-4}{x^{2}+1} d x= \) (Don't forget to include the constant of antidifferentiation \( c \) )

1 Answer

5 votes

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.

User Madars Vi
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories