How do you use partial fraction decomposition to decompose the fraction to integrate #(7)/(x^2+13x+40)#?

Answer 1

The integral equals #7/3ln|(x + 5)/(x +8)| + C#

We wish to find factors in the denominator. The trick is to find two numbers that multiply to #40# and add to #13#. Clearly these will be #8# and #5#.
#I = int 7/((x+ 5)(x+ 8))dx#

Now we can decompose in partial fractions.

#A/(x+ 5) + B/(x +8) = 7/((x +5)(x + 8))#
#A(x + 8) + B(x + 5) = 7#
#Ax + 8A + Bx + 5B = 7#
#(A + B)x + (8A + 5B) = 7#

Now we have a system of equations.

#{(A + B = 0), (8A + 5B = 7):}#

Substituting the first equation into the second we see that

#8A + 5(-A) = 7#
#3A = 7#
#A = 7/3#
Now clearly #B = -7/3# because #A+ B =0#. The integral becomes:
#I = int7/(3(x + 5)) - 7/(3(x + 8)) dx#
#I= 7/3ln|x +5| - 7/3ln|x + 8| + C#
#I = 7/3ln|(x + 5)/(x +8)| + C#

Hopefully this helps!

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Answer 2

To decompose the fraction (7)/(x^2 + 13x + 40) using partial fraction decomposition, follow these steps:

  1. Factor the denominator: x^2 + 13x + 40 = (x + 8)(x + 5).

  2. Write the fraction as a sum of two simpler fractions with undetermined coefficients: (7)/(x^2 + 13x + 40) = A/(x + 8) + B/(x + 5).

  3. Multiply both sides of the equation by (x + 8)(x + 5) to clear the denominators: 7 = A(x + 5) + B(x + 8).

  4. Expand and collect like terms: 7 = (A + B)x + (5A + 8B).

  5. Equate coefficients of like terms on both sides of the equation: For x terms: A + B = 0. For constant terms: 5A + 8B = 7.

  6. Solve the system of equations to find the values of A and B.

  7. Once you find the values of A and B, substitute them back into the partial fraction decomposition.

  8. Now integrate each term separately.

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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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