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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
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Use partial fraction decompostion to get::
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= [(A/(x-5))+(B/(x-3)] = (9x/4)/(x^2-7x+15)
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A(x-3)+B(x-5) = 9x+4
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A+B = 9
-3A-5B = 4
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3A+3B = 27
-3A-5B = 4
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-2B = 31
B = -31/2
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A = 49/2
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integra[((49/2)/(x-5) - (31/2)/(x-3)]
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= (49/2)ln(x-5) - (31/2)ln(x-3)
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Check the following site to see how this all works.
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http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/
partialfracsoldirectory/PartialFracSol.html#SOLUTION 1
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Cheers,
Stan H.
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