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

int(3-x)/(3x^2-2x-1)dx
Let's use partial fraction decomposition on the integrand,
(3-x)/(3x^2-2x-1)=(3-x)/(3x^2+x-3x-1)
=(3-x)/(x(3x+1)-1(3x+1))
=(3-x)/((3x+1)(x-1))
Now form the partial fractions using the denominator,
(3-x)/((3x+1)(x-1))=A/(3x+1)+B/(x-1)
Multiply equation by the denominator (3x+1)(x-1)
=>(3-x)=A(x-1)+B(3x+1)
=>3-x=Ax-A+3Bx+B
=>3-x=(A+3B)x+(-A+B)
comparing the coefficients of the like terms,
A+3B=-1 ----------------(1)
-A+B=3 ----------------(2)
Now let's solve the above equations to get A and B,
Add the equations 1 and 2,
4B=-1+3
4B=2
B=2/4
B=1/2
Plug in the value of B in equation 1,
A+3(1/2)=-1
A+3/2=-1
A=-1-3/2
A=-5/2
Plug in the value of A and B in the partial fraction template,
=(-5/2)/(3x+1)+(1/2)/(x-1)
=-5/(2(3x+1))+1/(2(x-1))
So, int(3-x)/(3x^2-2x-1)dx=int(-5/(2(3x+1))+1/(2(x-1)))dx
Apply the sum rule,
=int-5/(2(3x+1))dx+int1/(2(x-1))dx
Take the constant out,
=-5/2int1/(3x+1)dx+1/2int1/(x-1)dx
Now let's evaluate both the above integrals separately,
int1/(3x+1)dx
Apply integral substitution:u=3x+1
=>du=3dx
=int1/u(du)/3
Take the constant out,
=1/3int1/udu
Use the common integral:int1/xdx=ln|x|
=1/3ln|u|
Substitute back u=3x+1
=1/3ln|3x+1|
Now evaluate the second integral.
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|
int(3-x)/(3x^2-2x-1)dx=-5/2(1/3ln|3x+1|)+1/2ln|x-1|
Simplify and add a constant C to the solution,
=-5/6ln|3x+1|+1/2ln|x-1|+C