How do you integrate #(2x-1)/((x+1)(x-2)(x+3))# using partial fractions?
Before integration, partial fractions can be done as explained below
The partial fractions would thus be
Integration is now simple =
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To integrate ( \frac{2x - 1}{(x + 1)(x - 2)(x + 3)} ) using partial fractions, follow these steps:
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First, express the rational function as the sum of partial fractions: [ \frac{2x - 1}{(x + 1)(x - 2)(x + 3)} = \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{x + 3} ]
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Multiply both sides by the denominator ( (x + 1)(x - 2)(x + 3) ) to clear the fractions: [ 2x - 1 = A(x - 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x - 2) ]
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Expand and equate coefficients of like terms. This will give you a system of linear equations to solve for ( A ), ( B ), and ( C ).
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Once you find the values of ( A ), ( B ), and ( C ), rewrite the original function using these partial fractions.
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Integrate each partial fraction separately.
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Finally, add the integrated partial fractions together to obtain the final result.
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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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