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

1 Answer
Aug 11, 2016

#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#