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

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Answer 1 · 2017-02-25T07:37:33

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.

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