SOLUTION: integral 9x+4
----- dx
x2-7x+15
Algebra.Com
Question 556199: integral 9x+4
----- dx
x2-7x+15
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
integral [(9x+4)/(x^2-7x+15)] dx
---
Use partial fraction decompostion to get::
------------
= [(A/(x-5))+(B/(x-3)] = (9x/4)/(x^2-7x+15)
-----
A(x-3)+B(x-5) = 9x+4
----
A+B = 9
-3A-5B = 4
-----
3A+3B = 27
-3A-5B = 4
--------
-2B = 31
B = -31/2
---
A = 49/2
---------------
integra[((49/2)/(x-5) - (31/2)/(x-3)]
----
= (49/2)ln(x-5) - (31/2)ln(x-3)
----
Check the following site to see how this all works.
===============
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/
partialfracsoldirectory/PartialFracSol.html#SOLUTION 1
=====================
Cheers,
Stan H.
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