Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 18

For the given integral problem: int (6x)/(x^3-8)dx , we may partial fraction decomposition to expand the integrand: f(x)=(6x)/(x^3-8) .
The pattern on setting up partial fractions will depend on the factors of the denominator. For the given problem, the denominator is in a form of difference of perfect cube : x^3 -y^3 = (x-y)(x^2+xy+y^2)
Applying the special factoring on (x^3-8) , we get:
(x^3-8) =(x^3-2^3)
=(x-2)(x^2+x*2+2^2)
=(x-2)(x^2+2x+4)
For the linear factor (x-2) , we will have partial fraction: A/(x-2) .
For the quadratic factor (x^2+2x+4) , we will have partial fraction: (Bx+C)/(x^2+2x+4) .
The integrand becomes:
(6x)/(x^3-8) =A/(x-2) +(Bx+C)/(x^2+2x+4)
Multiply both side by the LCD =(x-2)(x^2+2x+4) :
((6x)/(x^3-8))*(x-2)(x^2+2x+4) =[ A/(x-2) +(Bx+C)/(x^2+2x+4)] *(x-2)(x^2+2x+4)
6x =A(x^2+2x+4) +(Bx+C)(x-2)
We apply zero-factor property on (x-2)(x^2+2x+4) to solve for values we can assign on x.
x-2 = 0 then x=2
x^2+2x+4=0 then x = -1+-sqrt(3)i
To solve for A , we plug-in x=2 :
6*2 =A(2^2+2*2+4) +(B*2+C)(2-2)
12 =A(4+4+4) +(2B+C)(0)
12 = 12A +0
12/12 = (12A)/12
A =1
To solve for C , plug-in A=1 and x=0 so that B*x becomes 0 :
6*0 =A(0^2+2*0+4) +(B*0+C)(0-2)
0 =1(0+0+4) +(0+C)(-2)
0=4 -2C
2C =4
(2C)/2=4/2
C=2
To solve for B , plug-in A=1 , C=2 , and x=1 :
6*1 =1(1^2+2*1+4) +(B*1+2)(1-2)
6 = 1+2+4 +(B+2)*(-1)
6 = 1+2+4 -B-2
6 = 5-B
6-5 =-B
1=-B
then B =-1
Plug-in A = 1 , B =-1, and C=2 , we get the partial fraction decomposition:
int (6x)/(x^3-8) dx = int [ 1/(x-2) +(-x+2)/(x^2+2x+4)] dx
=int [ 1/(x-2) -x/(x^2+2x+4)+2/(x^2+2x+4)] dx
Apply the basic integration property: int (u+-v+-w) dx = int (u) dx +- int (v) dx+- int (w) dx .
int [ 1/(x-2) -x/(x^2+2x+4)+2/(x^2+2x+4)] dx =int 1/(x-2) dx- int x/(x^2+2x+4)dx+ int 2/(x^2+2x+4) dx
For the first integral, we apply integration formula for logarithm: int 1/u du = ln|u|+C .
Let u =x-2 then du = dx
int 1/(x-2) dx =int 1/u du
= ln|u|
= ln|x-2|
For the second integral, we apply indefinite integration formula for rational function:
int x/(ax^2+bx+c) dx =1/(2a)ln|ax^2+bx+c| -b/(asqrt(4ac-b^2))arctan((2ax+b)/sqrt(4ac-b^2))
By comparing "ax^2 +bx +c " with "x^2+2x+4 ", we determine the corresponding values: a=1 , b=2 , and c=4 .
int x/(x^2+2x+4)dx=1/(2*1)ln|1x^2+2x+4| -2/(1sqrt(4*1*4-2^2))arctan((2*1x+2)/sqrt(4*1*4-2^2))
=1/2ln|x^2+2x+4|-2/sqrt(16-4)arctan((2x+2)/sqrt(16-4))
=1/2ln|x^2+2x+4|-2/sqrt(12)arctan((2x+2)/sqrt(12))
=1/2ln|x^2+2x+4|-2/(2sqrt(3))arctan((2(x+1))/(2sqrt(3)))
=1/2ln|x^2+2x+4| -1/sqrt(3)arctan((x+1)/sqrt(3))
=(ln|x^2+2x+4|)/2 -(arctan((x+1)/sqrt(3)))/sqrt(3)
Apply indefinite integration formula for rational function with a=1 , b=2 , and c=4 :
int 1/(ax^2+bx+c) dx = 2/sqrt(4ac-b^2)arctan((2ax+b)/sqrt(4ac-b^2)) +C
Then,
int 2/(x^2+2x+4) dx =2int 1/(x^2+2x+4) dx
=2*[2/sqrt(4*1*4-2^2)arctan((2*1x+2)/sqrt(4*1*4-2^2))]
= 2*[2/sqrt(16-4)arctan((2x+2)/sqrt(16-4))]
= 2*[2/(2sqrt(12))arctan((2x+2)/sqrt(12)) ]
= 2*[2/(2sqrt(3))arctan((2(x+1))/(2sqrt(3)))]
= 2*[1/sqrt(3)arctan((x+1)/sqrt(3))]
=2/sqrt(3)arctan((x+1)/sqrt(3))
=(2arctan((x+1)/sqrt(3)))/sqrt(3)
Combining the results, we get the indefinite integral as:
int (6x)/(x^3-8) dx =ln|x-2| - [(ln|x^2+2x+4|)/2 -arctan((x+1)/sqrt(3))/sqrt(3)]+(2arctan((x+1)/sqrt(3)))/sqrt(3) +C
=ln|x-2| -(ln|x^2+2x+4|)/2 +(arctan((x+1)/sqrt(3)))/sqrt(3)+(2arctan((x+1)/sqrt(3)) )/sqrt(3)+C
= (2ln|x-2|-ln|x^2+2x+4|)/2 +(arctan((x+1)/sqrt(3))+2arctan((x+1)/sqrt(3)))/sqrt(3) +C
= (ln|(x-2)^2/(x^2+2x+4)|)/2+(3arctan((x+1)/sqrt(3)))/sqrt(3) +C
= (ln|(x^2-4x+4)/(x^2+2x+4)|)/2 +sqrt(3)arctan((sqrt(3)(x+1))/3)+C
= (ln|(x^2-4x+4)/(x^2+2x+4)|)/2 +sqrt(3)arctan((xsqrt(3)+sqrt(3))/3)+C