Question

How do you integrate `int (4x-2) /( 3(x-1)^2)` using partial fractions?

Answer 1

`int(4x-2)/(3(x-1)^2)dx=4/3ln(x-1)-4/(3(x-1))+c`

Explanation:

Let us first find partial fractions of `(4x-2)/(3(x-1)^2)` and for this let

`(4x-2)/((x-1)^2)hArrA/(x-1)+B/(x-1)^2` or

`(4x-2)/((x-1)^2)hArr(A(x-1)+B)/((x-1)^2)=(Ax+B-A)/((x-1)^2)`

Hence `A=4` and `B-A=-2` i.e. `B=4-2=2`

Hence `int(4x-2)/(3(x-1)^2)dx=1/3int[2/(x-1)+2/(x-1)^2]dx`

= `2/3int2/(x-1)dx+2/3int2/(x-1)^2dx+k`

= `4/3ln(x-1)-4/(3(x-1))+c`