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\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


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\,