int (2x-5)/(x^2-4x+5)∫2x−5x2−4x+5
2x-5=2x-4-12x−5=2x−4−1
int (2x-5)/(x^2-4x+5)=int (2x-4-1)/(x^2-4x+5)d x∫2x−5x2−4x+5=∫2x−4−1x2−4x+5dx
"split the integration"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) ∫2x−5x2−4x+5=∫2x−4x2−4x+5dx−∫1x2−4x+5dx
color(red)(int (2x-4)/(x^2-4x+5)d x)∫2x−4x2−4x+5dx
"Substitute "u=x^2-4x+5" ; " d u=2x-4Substitute u=x2−4x+5 ; du=2x−4
color(red)(int (2x-4)/(x^2-4x+5)d x)=int (d u)/u=l n u∫2x−4x2−4x+5dx=∫duu=lnu
"Undo substitution"Undo substitution
color(red)(int (2x-4)/(x^2-4x+5)d x)=l n(|x^2-4x+5|)∫2x−4x2−4x+5dx=ln(∣∣x2−4x+5∣∣)
color(green)(int1/(x^2-4x+5)d x)=∫1x2−4x+5dx=
x^2-4x+5=(x-2)^2+1x2−4x+5=(x−2)2+1
"please remember that "int (d x)/(x^2+a^2)=1/a arc tan (x/a)+Cplease remember that ∫dxx2+a2=1aarctan(xa)+C
color(green)(int1/(x^2-4x+5)d x)=int (d x)/((x-2)^2+1)=arc tan (x-2)∫1x2−4x+5dx=∫dx(x−2)2+1=arctan(x−2)
int (2x-5)/(x^2-4x+5)=l n(|x^2-4x+5|-arc tan(x-2))+C∫2x−5x2−4x+5=ln(∣∣x2−4x+5∣∣−arctan(x−2))+C
