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How do you integrate #7/(2x-3)# using partial fractions?

Answer 1

Charles Anctil

You don't need to

This is already in a partial fraction form

#int 7/(2x-3) dx#

#7/2 int 2/(2x-3) dx#

#7/2 ln|2x-3|#

You only need to use partial fractions when there are multiple (x-k) (or quadratic) terms in the denominator, or the expression is factorable.

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Answer 2

Emily Ammerman

To integrate ( \frac{7}{2x - 3} ) using partial fractions, first express it as the sum of two fractions with unknown constants:

[ \frac{7}{2x - 3} = \frac{A}{2x - 3} ]

To find ( A ), multiply both sides by ( 2x - 3 ):

[ 7 = A(2x - 3) ]

Solve for ( A ):

[ A = \frac{7}{2x - 3} ]

Now, integrate both sides:

[ \int \frac{7}{2x - 3} , dx = \int \frac{A}{2x - 3} , dx ]

[ \int \frac{7}{2x - 3} , dx = \int \frac{7}{2x - 3} , dx ]

[ = \int A , dx ]

[ = A \int dx ]

[ = A x + C ]

Where ( C ) is the constant of integration. So the integral of ( \frac{7}{2x - 3} ) using partial fractions is:

[ \int \frac{7}{2x - 3} , dx = \frac{7x}{2} + 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.