Question

Question #36feb

Answer 1

`int (2x-3)/(x-1) = 2x-ln abs(x-1)+C`

Explanation:

Write the integrand function as:

`(2x-3)/(x-1) = 2( (x-3/2)/(x-1)) = 2( (x-1-1/2)/(x-1)) = 2 - 1/(x-1)`

So:

`int (2x-3)/(x-1) = int (2 - 1/(x-1))dx`

and using the linearity of integrals:

`int (2x-3)/(x-1) = 2 int dx - int dx/(x-1) = 2x-ln abs(x-1)+C`