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The answer in the back of the book is | The answer in the back of the book is | ||
| − | <math>\frac{1}{7}\ln{ | + | <math>\frac{1}{7}\ln{|(x+6)^2(x-1)^5|}+C</math> |
Revision as of 13:03, 13 October 2008
- I didn't get the right answer according to the back of the book. I got:
$ x + 4 = Ax + A + Bx - 6B = (A + B)x + A - 6B $
Meaning:
$ A + B = 1 $
and
$ A - 6B = 4 $
So, $ A = \frac{10}{7} $ and $ B = -\frac{3}{7} $
The answer in the back of the book is
$ \frac{1}{7}\ln{|(x+6)^2(x-1)^5|}+C $
