Now, letquot;s notice that we can use partial fractions on the series term to get, 1 1 1 1 = = - 2 i + 3i + 2 ( i + 2 )( i + 1) i + 1 i + 2...

Now, let"s notice that we can use partial fractions on the series term to get, 1 1 1 1 = = - 2 i + 3i + 2 ( i + 2 )( i + 1) i + 1 i + 2 I"ll leave the details of the partial fractions to you. By now you should be fairly adept at this since we spent a fair amount of time doing partial fractions back in the Integration Techniques chapter. If you need a refresher you should go back and review that section. So, what does this do for us?