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`
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