How do you integrate #(x+2)/(2x^3-8x)# using partial fractions?

Answer 1

# int \ (x+2)/(2x^3-8x) \ dx = 1/4ln|(x-2)/x| + C#

We seek:

# I = int \ (x+2)/(2x^3-8x) \ dx#

We can write as:

# I = int \ (x+2)/(2x(x^2-4)) \ dx# # \ \ = 1/2 \ int \ (x+2)/(x(x+2)(x-2) \ dx# (removable discontinuity) # \ \ = 1/2 \ int \ 1/(x(x-2) \ dx#

We can now decompose the integrand in to partial fractions:

# 1/(x(x-2)) -= A/x + B/(x-2) # # " " = (A(x-2)+Bx)/(x(x-2)) #

Leading to the identity:

# 1 -= A(x-2)+Bx #
Where #A,B# are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:
Put # x = 0 => 1=-2A => A = -1/2# Put # x = 2 => 1=2B => B = 1/2#

So we can now write:

# I = 1/2 \ int \ (-1/2)/x + (1/2)/(x-2) \ dx# # \ \ = 1/4 \ int \ 1/(x-2) - 1/x \ dx#

Which now consists of standard integral so we integrate to get:

# I = 1/4{ln|x-2| - ln|x| } + C# # \ \ = 1/4ln|(x-2)/x| + C#
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Answer 2

To integrate the expression (x+2)/(2x^3 - 8x) using partial fractions, follow these steps:

  1. Factor the denominator: 2x^3 - 8x = 2x(x^2 - 4) = 2x(x - 2)(x + 2).
  2. Express the fraction as a sum of partial fractions: (x+2)/(2x^3-8x) = A/x + B/(x-2) + C/(x+2).
  3. Multiply both sides of the equation by the denominator to clear the fractions: (x + 2) = A(x - 2)(x + 2) + Bx(x + 2) + Cx(x - 2).
  4. Solve for A, B, and C by equating coefficients of like terms.
  5. Once you find the values of A, B, and C, rewrite the original fraction with these values.
  6. Now, integrate each term separately.
  7. Finally, combine the integrals and simplify the result.

The specific values of A, B, and C can be determined by equating coefficients or by using other appropriate methods.

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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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