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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 -

    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

    = 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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