Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 6

int 2/(9x^2-1)
To solve using partial fraction method, the denominator of the integrand should be factored.
2/(9x^2-1) = 2/((3x-1)(3x+1))
Then, express it as sum of fractions.
2/((3x-1)(3x+1))=A/(3x-1)+B/(3x+1)
To solve for the values of A and B, multiply both sides by the LCD of the fractions present.
(3x-1)(3x+1)*2/((3x-1)(3x+1))=(A/(3x-1)+B/(3x+1))*(3x-1)(3x+1)
2 = A(3x+1) + B(3x-1)
Then, assign values to x in which either (3x+1) or (3x-1) will become zero.
So plug-in x=1/3 to get the value of A.
2=A(3*1/3+1) +B(3*1/3-1)
2=A(1+1) + B(1-1)
2=A(2) + B(0)
2=2A
1=A
Also, plug-in x=-1/3 to get the value of B.
2=A(3*(-1/3)+1)+B(3*(-1/3)-1)
2=A(-1+1)+B(-1-1)
2=A(0) + B(-2)
2=-2B
-1=B
So the partial fraction decomposition of the integrand is
int 2/(9x^2-1)dx
= int (2/((3x-1)(3x+1))dx
= int (1/(3x-1)-1/(3x+1))dx
Then, express it as difference of two integrals.
= int 1/(3x-1)dx - int 1/(3x+1)dx
To evaluate each integral, apply substitution method.

u=3x-1 du=3dx 1/3du=dx w=3x+1 dw=3dx 1/3dw=dx

Expressing the two integrals in terms of u and w, it becomes:
=int 1/u * 1/3du - int 1/w*1/3dw
=1/3 int 1/u du - 1/3int 1/w dw
To take the integral of this, apply the formula int 1/x dx = ln|x|+C .
=1/3ln|u| - 1/3ln|w| + C
And, substitute back u=3x-1 and w=3x+1.
=1/3ln|3x-1| -1/3ln|3x+1|+C

Therefore, int 2/(9x^2-1)=1/3ln|3x-1| -1/3ln|3x+1|+C .