How do you integrate #int (2x+1)/((x+4)(x-1)(x+7)) # using partial fractions?

Answer 1

#7/15ln|x+4| + 3/40ln|x-1| - 13/24ln|x+7| + c#

Since the factors on the denominator of the rational function are linear, then the numerators of the partial fractions will be constants, say A , B and C.

hence #(2x+1)/((x+4)(x-1)(x+7)) = A/(x+4) + B/(x-1) + C/(x+7) #

multiply through by (x+4)(x-1)(x+7) to obtain:

2x+1 = A(x-1)(x+7) + B(x+4)(x+7) + C(x+4)(x-1) ................(1)

The aim now is to find the value of A , B and C. Note that if x = -4 then the terms with B and C will be zero. If x = 1 the terms with A and C will be zero and if x =-7 the terms with A and B will be zero. This is the starting point in finding A , B and C.

let x = -4 in (1) : -7 = -15A #rArr A = 7/15 #
let x = 1 in (1) : 3 = 40B #rArr B = 3/40 #
let x =-7 in (1) : -13 = 24C #rArr C = -13/24 #
#(2x+1)/((x+4)(x-1)(x+7)) = (7/15)/(x+4) + (3/40)/(x-1) - (13/24)/(x+7)#

integral becomes:

#int(7/15)/(x+4) dx + int(3/40)/(x-1) dx - int(13/24)/(x+7) dx #
#= 7/15ln|x+4| + 3/40ln|x-1| - 13/24ln|x+7| + c #
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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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