I have trouble understanding what needs to be done here using the following explanation. Without solving this problem, or by solving something similar, could someone explain what needs to be done here with the following info.
Confirm that: $$\frac {1}{x^2 -1} = \frac {1}{2}(\frac {1}{x-1} - \frac {1}{x+1})$$
And use this equation to evaluate: $$\int_2^6 \frac {3}{x^2 -1}\,dx$$
On
Well lets start with the first equality:
$$\frac{1}{x^2-1}=\frac{1}{2}(\frac{1}{x-1}-\frac{1}{x+1})$$ Looking at the RHS: $$\frac{1}{2}(\frac{1}{x-1}-\frac{1}{x+1})=\frac{1}{2}(\frac{x+1}{x^2-1}-\frac{x-1}{x^2+1})$$ $$=\frac{1}{2}\frac{2}{x^2+1}=\frac{1}{x^2-1}$$ So that formula holds. And it comes from partial fraction decomposition. We can rewrite the LHS as follows: $$\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$$ $$1=A(x+1)+B(x-1)$$ Letting $x=-1$: $$1=-2B\Rightarrow B=\frac{-1}{2}$$ Letting $x=1$: $$1=2A\Rightarrow A=\frac{1}{2}$$ Thus we obtain: $$\frac{1}{x^2-1}=\frac{1}{2}\frac{1}{x-1}-\frac{1}{2}\frac{1}{x+1}$$ So we can use this formula to now evaluate the integral: $$\int_2^6\frac{3}{x^2-1}=3\int_2^6\frac{1}{x^2-1}$$ $$=\frac{3}{2}\int_2^6\frac{1}{x-1}-\frac{1}{x+1}$$ $$=\frac{3}{2}[\ln|x-1|-\ln|x+1|]|_2^6$$ $$=\frac{3}{2}[\ln|\frac{x-1}{x+1}|]|_2^6$$ $$=\frac{3}{2}(\ln(\frac{5}{7})-\ln(\frac{1}{3}))$$
Once you have confirmed that the first equation is valid, you can then use it:
\begin{align} \int_2^6\frac3{x^2-1}\,dx&=3\int_2^6\frac1{x^2-1}\,dx\\ &=3\int_2^6\frac12\left(\frac1{x-1}-\frac1{x+1}\right)\,dx\\ &=\frac32\int_2^6\frac1{x-1}\,dx-\frac32\int_2^6\frac1{x+1}\,dx \end{align}
As for how to evaluate these integrals, use the fact that $$\int_a^b\frac1x\,dx=\log|x|\bigg|_a^b.$$
To write it so the answer is in plain sight, we can let $u=x-1\implies du=dx$ and $v=x+1\implies dv=dx$ to produce the following: $$\frac32\int_2^6\frac1{x-1}\,dx-\frac32\int_2^6\frac1{x+1}\,dx=\frac32\int_1^5\frac1u\,du-\frac32\int_3^7\frac1v\,dv$$ and you should be able to take it from here.