# How do you integrate #int (3x + 2) / [(x - 1)(x + 4)] # using partial fractions?

We possess that

Hence

#int (3x+2)/[(x-1)(x+4)]dx=int 2/(x+4)dx+int 1/(x-1)dx= 2*logabs(x+4)+logabs(x-1)+c#

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To integrate ( \frac{3x + 2}{(x - 1)(x + 4)} ) using partial fractions, follow these steps:

- Factor the denominator: ( (x - 1)(x + 4) ).
- Decompose the fraction into partial fractions with unknown constants: ( \frac{A}{x - 1} + \frac{B}{x + 4} ).
- Multiply both sides of the equation by the denominator to clear the fraction.
- Substitute values of ( x ) that make each denominator zero to solve for the unknown constants ( A ) and ( B ).
- After finding the values of ( A ) and ( B ), rewrite the original fraction as the sum of the partial fractions.
- Integrate each partial fraction separately.
- Combine the results to get the final integral expression.

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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.

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