# Calc1.15

$\int {\frac {1}{x^{2}-1}}\,dx\,$

The key to this problem is realizing that we can split up ${\frac {1}{x^{2}-1}}\,$ into two fractions. This method is the method of Partial Fractions. We know

${\frac {1}{x^{2}-1}}={\frac {1}{(x+1)(x-1)}}\,$

Now, it is known that we can split this up into two fractions, one with a base of $x+1\,$ and the other with base $x-1\,$. Let

${\frac {1}{(x+1)(x-1)}}={\frac {A}{x+1}}+{\frac {B}{x-1}}\,$

Multiplying to get a common denominator and equating numerators gives us

$1=A(x-1)+B(x+1)\,$

Here, notice that this decomposition must be true for all $x\,$ so we can solve for $A\,$ and $B\,$ easily by picking values for $x\,$.

Let $x=1\,$, then $1=A(0)+B(2)\,$ which gives $B={\frac {1}{2}}\,$
Let $x=-1\,$, then $1=A(-2)+B(0)\,$ which gives $A=-{\frac {1}{2}}\,$

So our integral becomes

$\int {\frac {1}{x^{2}-1}}\,dx=\int \left[{\frac {-{\frac {1}{2}}}{x+1}}+{\frac {{\frac {1}{2}}}{x-1}}\right]\,dx\,$

This is now an integral for which we have a formula.

$\int \left[{\frac {-{\frac {1}{2}}}{x+1}}+{\frac {{\frac {1}{2}}}{x-1}}\right]\,dx=-{\frac {1}{2}}\ln |x+1|+{\frac {1}{2}}\ln |x-1|+C\,$