Question

How do you integrate `int (2x-5) / (x^2 -4x+ 5)` using partial fractions?

Answer 1

`int (2x-5)/(x^2-4x+5)=l n(|x^2-4x+5|-arc tan(x-2))+C`

Explanation:

`int (2x-5)/(x^2-4x+5)`

`2x-5=2x-4-1`

`int (2x-5)/(x^2-4x+5)=int (2x-4-1)/(x^2-4x+5)d x`

`"split the integration"`

`int (2x-5)/(x^2-4x+5)=color(red)(int (2x-4)/(x^2-4x+5)d x)-color(green)(int1/(x^2-4x+5)d x)`

`color(red)(int (2x-4)/(x^2-4x+5)d x)`

`"Substitute "u=x^2-4x+5" ; " d u=2x-4`

`color(red)(int (2x-4)/(x^2-4x+5)d x)=int (d u)/u=l n u`

`"Undo substitution"`

`color(red)(int (2x-4)/(x^2-4x+5)d x)=l n(|x^2-4x+5|)`

`color(green)(int1/(x^2-4x+5)d x)=`

`x^2-4x+5=(x-2)^2+1`

`"please remember that "int (d x)/(x^2+a^2)=1/a arc tan (x/a)+C`

`color(green)(int1/(x^2-4x+5)d x)=int (d x)/((x-2)^2+1)=arc tan (x-2)`

`int (2x-5)/(x^2-4x+5)=l n(|x^2-4x+5|-arc tan(x-2))+C`