How do you integrate #f(x)=(x^2-2x)/((x^2-3)(x-2)(x+8))# using partial fractions?

Here,

#I=int(x^2-2x)/((x^2-3)(x-2)(x+8))dx#

#=int(x(x-2))/((x^2-3)(x-2)(x+8))dx#

#I=intx/((x^2-3)(x+8))dx#

Partial Fractions :

#x/((x^2-3)(x+8))=A/(x+8)+(Bx+C)/(x^2-3)#

#=>x=A(x^2-3)+(Bx+C)(x+8)#

#=>x=Ax^2-3A+Bx^2+8Bx+Cx+8C#

#=>x=x^2(A+B)+x(8B+C)+8C-3A#

Comparing coefficient of #x^2, x and #constant term :

#A+B=0to(1),8B+C=1 to(2), 8C-3A=0 to(3)#

From#(1)# , #B=-Ato(4) and# from #(3)# , #C=(3A)/8to(5)#

So, from #(2)# , #-8A+(3A)/8=1=>-(61A)=8=>A=-8/61#

From #(4)# , #B=8/61#

From #(5)# ,#C=3/8(-8/61)=>C=-3/61#

Hence,

#I=int[(-8/61)/(x+8)+((8/61)x-(3/61))/(x^2-3)]dx#

#=-8/61int1/(x+8)dx+8/61intx/(x^2-3)dx-3/61int1/(x^2-3)dx#

#=-8/61ln|x+8|+4/61color(red)(int(2x)/(x^2-3)dx)-3/61color(blue)(int1/(x^2- (sqrt3)^2)dx#

=#-8/61ln|x+8|+4/61color(red)(ln|x^2-3|)-3/61*color(blue)(1/(2sqrt3)ln|(x- sqrt3)/(x+sqrt3)|+C#

#=1/61{4ln|x^2-3|-8ln|x+8|-sqrt3/2ln|(x-sqrt3)/(x+sqrt3)|}+C#

Note:

#color(red)((1)int(f'(x))/(f(x))dx=ln|f(x)|+c#

#color(blue)((2)int1/(X^2-A^2)dX=1/(2A)ln|(X-A)/(X+A)|+c#

The function is

#f(x)=(x^2-2x)/((x^2-3)(x-2)(x+8))#

#=(xcancel(x-2))/((x^2-3)cancel(x-2)(x+8))#

#=x/((x^2-3)(x+8))#

Perform the decomposition into partial fractions

#x/((x^2-3)(x+8))=(Ax+B)/(x^2-3)+C/(x+8)#

#=((Ax+B)(x+8)+(C)(x^2-3))/((x^2-3)(x+8))#

The denominators are the same, compare the numerators

#x=(Ax+B)(x+8)+(C)(x^2-3)#

Compare the LHS and the RHS

Coefficients of #x^2#, #=>#, #0=A+C#, #=>#, #A=-C#

Let #x=-8#, #=>#, #-8=61C#, #=>#, #C=-8/61#

#A=8/61#

#0=8B-3C#, #=>#, #B=3/8*-8/61=-3/61#

Therefore,

#x/((x^2-3)(x+8))=(8/61x-3/61)/(x^2-3)+(-8/61)/(x+8)#

The integral is

#int(xdx)/((x^2-3)(x+8))=1/61int((8x-3)dx)/(x^2-3)-8/61int(dx)/(x+8)#

#I=I_1-I_2#

#I_2=8/61int(8dx)/(x+8)=8/61ln(x+8)#

For the calculation of #I_2#, perform the decomposition into partial fractions.

#(8x-3)/(x^2-3)=A/(x+sqrt3)+B/(x-sqrt3)#

#=(A(x-sqrt3)+B(x+sqrt3))/(x^2-3)#

Comparing the numerators,

#8x-3=A(x-sqrt3)+B(x+sqrt3)#

Let #x=-sqrt3#, #=>#, #-8sqrt3-3=A*-2sqrt3#, #=>#, #A=(8sqrt3+3)/(2sqrt3)=4+1/2sqrt3#

Let #x=sqrt3#, #=>#, #8sqrt3-3=B*2sqrt3#, #=>#, #B=(8sqrt3-3)/(2sqrt3)=4-1/2sqrt3#

Therefore,

#(8x-3)/(x^2-3)=(4+1/2sqrt3)/(x+sqrt3)+(4-1/2sqrt3)/(x-sqrt3)#

So,

#I_1=1/61int((8x-3)dx)/(x^2-3)=1/61int((4+1/2sqrt3)dx)/(x+sqrt3)+1/61int((4-1/2sqrt3)dx)/(x-sqrt3)#

#=(4+1/2sqrt3)/61ln(x+sqrt3)+(4-1/2sqrt3)/61ln(x-sqrt3)#

#=(8+sqrt3)/122ln(x+sqrt3)+(8-sqrt3)/122ln(x-sqrt3)#

And finally,

#int(xdx)/((x^2-3)(x+8))=(8+sqrt3)/122ln(|x+sqrt3|)+(8-sqrt3)/122ln(|x-sqrt3|)-8/61ln(|x+8|)+C#