Question

How do you integrate `int (5x)/((x-1)(x+3))` using partial fractions?

Answer 1

The answer is `=5/4ln(x-1)+15/4ln(x+3)+C`

Explanation:

First, let's do the decomposition into partial fractions,
`(5x)/((x-1)(x+3))=A/(x-1)+B/(x+3)`

`=((x+3)+B(x-1))/((x-1)(x+3))`

`:. 5x=A(x+3)+B(x-1)`
Let `x=-3` `=>` `-15=-4B` `=>` `B=15/4`

Let `x=1` `=>` `5=4A` `=>` `A=5/4`

`(5x)/((x-1)(x+3))=(5/4)/(x-1)+(15/4)/(x+3)`

So, `int(5xdx)/((x-1)(x+3))=int(5/4dx)/(x-1)+int(15/4dx)/(x+3)`

`=5/4ln(x-1)+15/4ln(x+3)+C`