Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 16

int(8x)/(x^3+x^2-x-1)dx
(8x)/(x^3+x^2-x-1)=(8x)/((x^3+x^2)-1(x+1))
=(8x)/((x^2(x+1)-1(x+1)))
=(8x)/((x+1)(x^2-1))
=(8x)/((x+1)(x+1)(x-1))
=(8x)/((x-1)(x+1)^2)
Now let's form the partial fraction template,
(8x)/((x-1)(x+1)^2)=A/(x-1)+B/(x+1)+C/(x+1)^2
Multiply the equation by the denominator,
8x=A(x+1)^2+B(x-1)(x+1)+C(x-1)
8x=A(x^2+2x+1)+B(x^2-1)+C(x-1)
8x=Ax^2+2Ax+A+Bx^2-B+Cx-C
8x=(A+B)x^2+(2A+C)x+A-B-C
Comparing the coefficients of the like terms,
A+B=0 -----------------(1)
2A+C=8 -----------------(2)
A-B-C=0 ---------------(3)
From equation 1,
B=-A
Substitute B in equation 3,
A-(-A)-C=0
2A-C=0 ---------------(4)
Now add equations 2 and 4,
4A=8
A=8/4
A=2
B=-A=-2
Plug in the value of A in equation 4,
2(2)-C=0
C=4
Plug in the values of A, B and C in the partial fraction template,
(8x)/((x-1)(x+1)^2)=2/(x-1)+(-2)/(x+1)+4/(x+1)^2
int(8x)/(x^3+x^2-x-1)dx=int(2/(x-1)-2/(x+1)+4/(x+1)^2)dx
Apply the sum rule,
=int2/(x-1)dx-int2/(x+1)dx+int4/(x+1)^2dx
Take the constant out,
=2int1/(x-1)dx-2int1/(x+1)dx+4int1/(x+1)^2dx
Now let's evaluate the above three integrals separately,
int1/(x-1)dx
Apply integral substitution:u=x-1
du=1dx
=int1/udu
use the common integral int1/xdx=ln|x|
=ln|u|
Substitute back u=x-1
=ln|x-1|
Now let's evaluate second integral,
int1/(x+1)dx
Apply integral substitution: u=x+1
du=dx
=int1/udu
=ln|u|
Substitute back u=x+1
=ln|x+1|
Now evaluate the third integral,
int1/(x+1)^2dx
apply integral substitution: u=(x+1)
du=dx
=int1/u^2du
=intu^(-2)du
apply power rule,
=u^(-2+1)/(-2+1)
=-1/u
Substitute back u=x+1
=-1/(x+1)
int(8x)/(x^3+x^2-x-1)dx=2ln|x-1|-2ln|x+1|+4(-1/(x+1))
Simplify and add a constant C to the solution,
=2ln|x-1|-2ln|x+1|-4/(x+1)+c

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