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

1 Answer
Elijah T.
Feb 25, 2017

You don't need to

Explanation:

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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