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

Question

William Abdalla · Accepted Answer

#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#

Christopher Allers · Answer

undefined To integrate $ \frac{6}{{(x-5)(x^2-2x+3)}} $ using partial fractions, first factor the denominator completely. $ x^2-2x+3 $ does not factor further, so it remains as is. Then, express $ \frac{6}{{(x-5)(x^2-2x+3)}} $ as the sum of two fractions with undetermined coefficients, $ \frac{A}{x-5} $ and $ \frac{Bx + C}{x^2-2x+3} $. Find the values of $ A $, $ B $, and $ C $ by equating the original expression with the partial fraction decomposition. After finding $ A $, $ B $, and $ C $, integrate each partial fraction separately. Finally, combine the integrals to get the result. The integration of $ \frac{A}{x-5} $ is $ A \ln|x-5| $, and the integration of $ \frac{Bx + C}{x^2-2x+3} $ involves completing the square and applying a trigonometric substitution.