# How do you integrate #int (x+13)/(x^3+2x^2-5x-6)# using partial fractions?

The answer is

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To integrate ( \frac{x + 13}{x^3 + 2x^2 - 5x - 6} ) using partial fractions:

- Factor the denominator ( x^3 + 2x^2 - 5x - 6 ) into linear factors.
- Decompose the rational function into partial fractions.
- Determine the unknown constants.
- Integrate each partial fraction term.
- Combine the integrated terms to find the final result.

The steps are as follows:

- Factor the denominator: ( x^3 + 2x^2 - 5x - 6 = (x + 3)(x - 1)(x + 2) ).
- Decompose the rational function into partial fractions: ( \frac{x + 13}{(x + 3)(x - 1)(x + 2)} = \frac{A}{x + 3} + \frac{B}{x - 1} + \frac{C}{x + 2} ).
- Determine the unknown constants ( A ), ( B ), and ( C ).
- Multiply both sides by the denominator to clear the fractions: ( x + 13 = A(x - 1)(x + 2) + B(x + 3)(x + 2) + C(x + 3)(x - 1) ).
- Substitute convenient values of ( x ) to solve for ( A ), ( B ), and ( C ).
- Integrate each partial fraction term separately.
- Combine the integrated terms to find the final result.

The integration of each partial fraction term will yield natural logarithmic functions and inverse tangent functions, depending on the constants ( A ), ( B ), and ( C ) obtained in step 5.

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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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