# integration using partial fractions

• Feb 3rd 2009, 11:48 PM
twilightstr
integration using partial fractions
integrate 1/(x^2 - 1) from 2 to 3.
A/(x+1) + B(x-1)
A(x-1) +B(x+1) A=-1/2 B=1/2
-1/2ln(x+1) + 1/2ln(x-1)]from 2 to 3.
= -1/2ln2^2 + 1/2ln2 - [-1/2ln3].
=-1/2ln2 - 1/2ln3 + C this is not right...the actual answer is 1/2ln(3/2).
• Feb 4th 2009, 12:46 AM
mylestone
Quote:

Originally Posted by twilightstr
integrate 1/(x^2 - 1) from 2 to 3.
A/(x+1) + B(x-1)
A(x-1) +B(x+1) A=-1/2 B=1/2
-1/2ln(x+1) + 1/2ln(x-1)]from 2 to 3.
= -1/2ln2^2 + 1/2ln2 - [-1/2ln3].
=-1/2ln2 - 1/2ln3 + C this is not right...the actual answer is 1/2ln(3/2).

Fantastic, let's take it from $- \frac{1}{2} ln(x+1) + \frac{1}{2} ln(x-1)$. I don't like an expression that leads-off with a negative value if I can help it, so let's rewrite this as $\frac{1}{2} ln(x-1) - \frac{1}{2} ln(x+1)$. Let's factor that irritating fraction out ('cause it's just going to screw with our heads anyways) so that we have $\frac{1}{2} (ln(x-1) - ln(x+1))$. That much you had for yourself already, all I did was some rearranging to clean things up, because now that we've prettied it up and pushed that fraction aside, we can focus on the neat part: the difference of logarithms. It just so happens that $ln(stuff) - ln(something) = ln(\frac{stuff}{something})$. Sweet! That should get you through, and remember that fact right there--it comes in handy. Oh, and I noticed that toward the end of your calculation you were still carrying a constant-of-integration "C". That constant should have been cancelled out by itself when you evaluated over the interval-of-interest (so you can effectively ignore these C's in situations like this).