How do you integrate #int (2x - 1) / (x^2 + 5x - 6) dx# using partial fractions?
The integral equals
So,
Put back into the original equation.
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To integrate ( \int \frac{2x - 1}{x^2 + 5x - 6} ) using partial fractions, we first factor the denominator: [ x^2 + 5x - 6 = (x - 1)(x + 6) ]
Now, we express the fraction in terms of partial fractions: [ \frac{2x - 1}{x^2 + 5x - 6} = \frac{A}{x - 1} + \frac{B}{x + 6} ]
Multiplying both sides by the denominator ( (x^2 + 5x - 6) ), we get: [ 2x - 1 = A(x + 6) + B(x - 1) ]
Expanding and equating coefficients, we find ( A ) and ( B ): [ 2x - 1 = Ax + 6A + Bx - B ] [ 2x - 1 = (A + B)x + (6A - B) ]
By comparing coefficients, we have: [ A + B = 2 ] [ 6A - B = -1 ]
Solving these equations simultaneously, we find ( A = \frac{1}{7} ) and ( B = \frac{13}{7} ).
Now, we can rewrite the original integral as: [ \int \frac{1}{7} \cdot \frac{1}{x - 1} + \frac{13}{7} \cdot \frac{1}{x + 6} , dx ]
Integrating each term separately, we get: [ \frac{1}{7} \ln|x - 1| + \frac{13}{7} \ln|x + 6| + C ]
Where ( C ) is the constant of integration. Thus, the integral of ( \frac{2x - 1}{x^2 + 5x - 6} ) using partial fractions is ( \frac{1}{7} \ln|x - 1| + \frac{13}{7} \ln|x + 6| + C ).
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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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