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March 27, 2015

Posted by **kchik** on Tuesday, December 4, 2012 at 12:47pm.

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 2:41pmI 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[(x-2)(x+1)]+B[(x(x+1)]+C[x(x-2)]

= Ax^2-2Ax-2A

+ Bx^2+Bx

+ Cx^2-2Cx

= (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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