Calc1.15

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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

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\,