Question

How do you integrate `(x-5)/(x-2)^2` using partial fractions?

Answer 1

`int (x-5)/(x-2)^2 .d x=l n(|x-2|)-3/(x-2)+C`

Explanation:

`int (x-5)/(x-2)^2 .d x=?`

`"let us write (x-5) as (x-2-3)"`

`int (x-2-3)/(x-2)^2 * d x`

`"split (x-2-3 )`

`int cancel((x-2))/cancel((x-2)^2)* d x-int 3/(x-2)^2 *d x`

`int (d x)/(x-2)-3 int (d x)/(x-2)^2`

`"1- solve " int (d x)/(x-2)`

`"substitute u=x-2" ; "d u=d x`

`int (d x)/(x-2)=(d u)/u=l n u`

`"undo substitution "`

`int (d x)/(x-2)=l n (x-2)`

`"2- now solve "3 int (d x)/(x-2)^2`

`u=x-2" ; " d x=d u" ; "3 int (d u)/u^2=3int u^-2 d u`

`3int (d x)/(x-2)=1/(-2+1 )u^(-2+1)=-3*u^(-1)=-3/u`

`"undo substitution"`

`3int(d x)/(x-2)^2=-3/(x-2)`

`"The problem is solved:"`

`int (x-5)/(x-2)^2 .d x=l n(|x-2|)-3/(x-2)+C`