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

Question

2234 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-11T16:12:58

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

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

$2x-5=2x-4-1$

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

$"split the integration"$

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

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

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

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

$"Undo substitution"$

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

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

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

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

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

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

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