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