How do you integrate #f(x)=1/((x-2)(x-5)(x+3))# using partial fractions?

Answer 1

#1/24ln|x-5| + 1/40ln|x+3| - 1/15ln|x-2| + c #

Since the factors on the denominator are linear then the numerators of the partial fractions will be constants , say A , B and C #rArr 1/((x-2)(x-5)(x+3)) = A/(x-2) + B/(x-5) + C/(x+3) #

now, multiply through by (x-2)(x-5)(x+3)

so 1 = A(x-5)(x+3) + B(x-2)(x+3) + C(x-2)(x-5) ................................(1)

We now have to find the values of A , B and C. Note that if x = 2 , the terms with B and C will be zero. If x = 5 , the terms with A and C will be zero and if x = -3, the terms with A and B will be zero. Making use of this fact , we obtain.

let x = 2 in (1) : 1 = -15A # rArr A = -1/15 #
let x = 5 in (1) : 1 = 24B #rArr B = 1/24 #
let x = -3 in (1) : 1 = 40C # rArr C = 1/40 #
#rArr 1/((x-2)(x-5)(x+3)) = (-1/15)/(x-2) + (1/24)/(x-5) + (1/40)/(x+3)#

Integral now becomes.

#-1/15intdx/(x-2) + 1/24intdx/(x-5) + 1/40intdx/(x+3) #
# = 1/24ln|x-5| + 1/40ln|x+3| - 1/15ln|x-2| + c #
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Answer 2

To integrate ( f(x) = \frac{1}{(x-2)(x-5)(x+3)} ) using partial fractions, follow these steps:

  1. Factor the denominator if necessary.
  2. Write ( f(x) ) as a sum of partial fractions.
  3. Determine the constants for each partial fraction.
  4. Integrate each partial fraction separately.
  5. Combine the results to find the final integral.

Here are the detailed steps:

  1. Factor the denominator: ( (x-2)(x-5)(x+3) ) cannot be further factored since it's a product of distinct linear factors.

  2. Write ( f(x) ) as partial fractions: ( f(x) = \frac{A}{x-2} + \frac{B}{x-5} + \frac{C}{x+3} )

  3. Determine the constants ( A ), ( B ), and ( C ): To find ( A ), ( B ), and ( C ), you can use methods like equating coefficients or substitution.

  4. Integrate each partial fraction separately:

    • ( \int \frac{A}{x-2} , dx = A \ln|x-2| + C_1 )
    • ( \int \frac{B}{x-5} , dx = B \ln|x-5| + C_2 )
    • ( \int \frac{C}{x+3} , dx = C \ln|x+3| + C_3 )
  5. Combine the results: ( \int f(x) , dx = A \ln|x-2| + B \ln|x-5| + C \ln|x+3| + C_1 + C_2 + C_3 )

Where ( A ), ( B ), and ( C ) are the constants determined in step 3, and ( C_1 ), ( C_2 ), and ( C_3 ) are integration constants.

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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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