# int 2/(9x^2-1) dx Use partial fractions to find the indefinite integral

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 .