How do you integrate #6/((x-5)(x^2-2x+3))# using partial fractions?

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Answer 1 · 2018-05-17T18:52:43

#I=1/3ln|x-5|-1/6ln|x^2-2x+3|-4/(3sqrt2)tan^-1((x-1)/sqrt2)#+c

Here,

#I=int6/((x-5)(x^2-2x+3))dx#

Partial Fractions :

#color(blue)(6/((x-5)(x^2-2x+3))=A/(x-5)+(Bx+C)/(x^2-2x+3)#

#=>6=A(x^2-2x+3)+Bx(x-5)+C(x-5)#

#=>6=x^2(A+B)+x(-2A-5B+C)+3A-5C#

Comparing Coefficient of x^2, x and constant term,

#A+B=0=>A=-Bto(1)#

#-2A-5B+C=0to(2) #

#3A-5C=6to(3)#

Subst. #A=-B# into #(2)#

#:.2B-5B+C=0=>-3B+C=0=>C=3B#

From #(3)# we get,

#3(-B)-5(3B)=6=>-18B=6=>B=-1/3#

From, (1) #A=-(-1/3)=1/3 and C=3(-1/3)=-1#

#i.e. color(blue)( A=1/3,B=-1/3,C=-1#

So

#I=int(1/3)/(x-5)dx+int((-1/3x)-1)/(x^2-2x+3)dx#

#=1/3int1/(x-5)dx-1/3int(x+3)/(x^2-2x+3)dx#

#=1/3ln|x-5|-1/3*1/2int(2x+6)/(x^2-2x+3)dx#

#=1/3ln|x-5|-1/6int(2x-2+8)/(x^2-2x+3)dx#

#=1/3ln|x-5|-1/6int(2x-2)/(x^2-2x+3)dx-1/6int8/(x^2-2x+3)dx#

#=1/3ln|x-5|-1/6int(d/(dx)(x^2-2x+3))/(x^2-2x+3)dx#

#color(white)(.........................)#.#-8/6int1/((x^2-2x+1+2))dx#

#=1/3ln|x-5|-1/6ln|x^2-2x+3|#
#color(white)(..............................)-4/3color(red)(int1/((x-1)^2+(sqrt2)^2)dx#

=#1/3ln|x-5|-1/6ln|x^2-2x+3|-4/3*color(red)(1/sqrt2 tan^-1((x-1)/sqrt2)#+ c

#I=1/3ln|x-5|-1/6ln|x^2-2x+3|-4/(3sqrt2)tan^-1((x-1)/sqrt2)#+c

Note :

#color(red)(int1/(x^2+a^2)dx=1/atan^-1(x/a)+c#