How do you integrate #int (2x - 1) / (x^2 + 5x - 6) dx# using partial fractions?

Answer 1

The integral equals # 13/7ln|x + 6| + 1/7ln|x - 1| + C#.

We can factor the denominator as #(x + 6)(x - 1)#.

So,

#A/(x + 6) + B/(x- 1) = (2x - 1)/((x + 6)(x - 1))#
#A(x - 1) + B(x + 6) = 2x - 1#
#Ax - A + Bx + 6B = 2x - 1#
#(A + B)x +(-A + 6B) = 2x - 1#
We can now write a systems of equations to solve for #A# and #B#.
#A + B = 2# #-A + 6B = -1#
#B = 2- a#
#-A + 6(2 - A) = -1#
#-A + 12 - 6A = -1#
#-7A = -13#
#A = 13/7#
#13/7 + B = 2#
#B = 2- 13/7#
#B = 1/7#

Put back into the original equation.

The integral become #int((13/7)/(x + 6) + (1/7)/(x - 1)) dx#
We now use the rule #int(1/x)(dx) = ln|x| + C# to get rid of the integral.
Letting the function be #f(x)#.
#f(x) = 13/7ln|x + 6| + 1/7ln|x - 1| + C#

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Answer 2

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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Answer from HIX Tutor

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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