Answer
$$\int\frac{2}{9x^2-1}dx=\frac{1}{3}\ln|3x-1|-\frac{1}{3}\ln|3x+1|+c$$
Work Step by Step
We can transform the integral into:
$$\int\frac{2}{9x^2-1}dx=\int\frac{2}{(3x)^2-1}dx=\int\frac{2}{(3x-1)(3x+1)}dx$$
Now we can decompose into partial fractions following the steps below:
Step 1: Since the denominator is already factored we decompose as:
$$\frac{2}{(3x-1)(3x+1)}=\frac{A}{3x-1}+\frac{B}{3x+1}$$
Step 2: Put the fractions back together with the common denominator:
$$\frac{2}{(3x-1)(3x+1)}=\frac{A}{3x-1}+\frac{B}{3x+1}=\frac{A(3x+1)+B(3x-1)}{(3x-1)(3x+1)}=\frac{3x(A+B)+A-B}{(3x-1)(3x+1)}$$
Step 3: Equate the numerators of the first and the final equation:
$$2=3x(A+B)+A-B$$
Step 4: Equate the coefficients multiplying the same powers of $x$ on the both side of the last equation. Since there is no $x$ on the left side we take $0$ for its' coefficient.
$A-B=2,3(A+B)=0\Rightarrow A-B=2,A+B=0$
From the second equation we get $A=-B$ and substituting that into the first one we get $2=A-B=A+A=2A\Rightarrow A=1$. Since $A=-B$ we have $B=-1.$
Now we have:
$$\frac{2}{(3x-1)(3x+1)}=\frac{A}{3x-1}+\frac{B}{3x+1}=\frac{1}{3x-1}-\frac{1}{3x+1}$$
We can go back to solving our integral:
$$\int\frac{2}{9x^2-1}dx=\int\frac{2}{(3x-1)(3x+1)}dx=\int\left(\frac{1}{3x-1}-\frac{1}{3x+1}\right)dx=\int\frac{1}{3x-1}dx-\frac{1}{3x+1}dx$$
We get two integrals we are going to solve separately:
Integral 1: To solve
$$\int\frac{1}{3x-1}dx$$
we will use substitution $3x-1=t$ which gives us $3dx=dt$. Putting this into the integral we have:
$$\int\frac{1}{3x-1}dx=\frac{1}{3}\int\frac{dt}{t}=\frac{1}{3}\ln|t|+c_1$$
where $c_1$ is arbitrary constant. Now we have to express our solution in terms of $x$ expressing $t$ in terms of $x$:
$$\int\frac{1}{3x-1}dx=\frac{1}{3}\ln|t|+c_1=\frac{1}{3}\ln|3x-1|+c_1$$
Integral 2:
To solve the integral
$$\int\frac{1}{3x+1}dx$$
we will use substitution $3x+1=t\Rightarrow 3dx=dt$. Substituting this into the integral we get:
$$\int\frac{1}{3x+1}dx=\frac{1}{3}\int\frac{dt}{t}=\frac{1}{3}\ln|t|+c_2,$$
where $c_2$ is arbitrary constant. Now express solution in terms of $x$ by expressing $t$ in terms of $x$:
$$\int\frac{1}{3x+1}dx=\frac{1}{3}\ln|t|+c_2=\frac{1}{3}\ln|3x+1|+c_2$$
Now we calculated both integrals we get:
$$\int\frac{2}{9x^2-1}dx=\int\frac{1}{3x-1}dx-\int\frac{1}{3x+1}dx=\frac{1}{3}\ln|3x-1|+c_1-\frac{1}{3}\ln|3x+1|-c_2=\frac{1}{3}\ln|3x-1|-\frac{1}{3}\ln|3x+1|+c$$
where we used that $c=c_1-c_2$ which is arbitrary constant as well.
