How do you find #int ( 3x-1)/(x^2+2x-8) dx# using partial fractions?

Answer 1

#(13ln abs(x+4)+ 5ln abs(x-2))/6#

#int (3x-1)/(x^2+2x-8)dx=int (3x-1)/((x+4)(x-2))dx#
To decompose the function into partial fractions, there are #A# and #B# such that:
#A/(x+4)+B/(x-2)=(3x-1)/((x+4)(x-2))#
#A(x-2)+B(x+4)=3x-1# #(A+B)x+(4B-2A)=3x-1#
Now solve for #A# and #B#. #A+B=3# #4B-2A=-1# #->A=13/6, B=5/6#
Integrate: #int (13/6)/(x+4)dx+int (5/6)/(x-2)dx#
#13/6 int1/(x+4)dx+5/6 int 1/(x-2)dx#
Set #u=x+4# in the first and #v=x-2# in the second so that both have derivative of #1#.
#13/6 int1/u du+5/6int1/v dv#
#13/6 lnu+5/6 ln v#
#13/6 ln abs(x+4)+ 5/6 ln abs(x-2)#
#(13ln abs(x+4)+ 5ln abs(x-2))/6#
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Answer 2

To find ( \int \frac{3x - 1}{x^2 + 2x - 8} , dx ) using partial fractions, follow these steps:

  1. Factor the denominator ( x^2 + 2x - 8 ) to identify its roots.
  2. Decompose the fraction into partial fractions with undetermined coefficients.
  3. Solve for the coefficients by equating the original expression with the decomposition.
  4. Integrate each partial fraction separately.
  5. Combine the integrals to find the final result.

Here's a step-by-step breakdown:

  1. Factor the denominator: [ x^2 + 2x - 8 = (x + 4)(x - 2) ]

  2. Decompose the fraction: [ \frac{3x - 1}{x^2 + 2x - 8} = \frac{A}{x + 4} + \frac{B}{x - 2} ]

  3. Solve for coefficients ( A ) and ( B ): [ 3x - 1 = A(x - 2) + B(x + 4) ]

  4. Equate coefficients: [ 3x - 1 = Ax - 2A + Bx + 4B ] [ (3 - A + B)x + (-2A + 4B) = 3x - 1 ]

Matching coefficients: [ 3 - A + B = 3 ] (for ( x )) [ -2A + 4B = -1 ] (constant term)

Solving these equations gives: [ A = 1 ] [ B = 2 ]

  1. Integrate each partial fraction: [ \int \frac{1}{x + 4} , dx = \ln|x + 4| + C_1 ] [ \int \frac{2}{x - 2} , dx = 2\ln|x - 2| + C_2 ]

  2. Combine the integrals: [ \int \frac{3x - 1}{x^2 + 2x - 8} , dx = \ln|x + 4| + 2\ln|x - 2| + C ]

Where ( C ) is the constant of integration.

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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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