Posted by kchik on Tuesday, December 4, 2012 at 12:47pm.
INTEGRATION USING PARTIAL FRACTIONS
a) 4x^2-3x+2 / x^3-x^2-2x dx
b) x^2 / (x+1)(x+1)^2 dx
c) 3-3x / 2x^2+6x dx
d) x^2+2x-1 / (x^2+1)(x-1) dx
##anybody can help me for this question?
calculus - Steve, Tuesday, December 4, 2012 at 2:41pm
I assume you have trouble doing the partial fractions, not the integration?
x^3-x^2-2x = x(x-2)(x+1)
so, what you are looking for is a combination of fractions
A/x + B/(x-2) + C/(x+1) which when placed over a common denominator of (x^3-x^2-2x) make a numerator of (4x^2-3x+2).
To add those simpler fractions, you have a numerator of
= (A+B+C)x^2 + (-2A+B-2C)x + (-2A)
for that to be identical to 4x^2-3x+2, you need
A+B+C = 4
-2A+B-2C = -3
-2A = 2
or, A = -1, B=3, C=2
so you have -1/x + 3/(x-2) + 2/(x+1)
integrate that to get some logs.
do the other likewise. Note:
quadratic factors below require (Ax+B) above
repeated factors below require all powers. e.g. */(x+1)^2 --> A/(x+1) + B/(x+1)^2, though some numerators may turn out to be zero.
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