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Calculus

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Evaluate the integral
INT(x^4+6x^3+5x^2+8)/(x^2+6x+5)dx

Please help.

  • Calculus -

    The integrand is:

    f(x) = (x^4+6x^3+5x^2+8)/(x^2+6x+5)

    The denominator can be factored as:

    x^2+6x+5 = (x+1)(x+5)

    This means that f(x) has singularities at x = -1 and x = -5, if the numerator isn't zero there. To find the partial fraction expansion, you can simply calculate the singular behavior near these points, subtract from f(x), leving you with a rational function without any singularities, which is therefore a polynomial (which you can easily find).

    This has the advantage of not having to solve equations to find the partial fraction expansion. To find the bahavior near x = -1, you expand the function in powers of (x+1), only the leading term is singular

    f(x) = 2/(x+1) + non-singular terms

    The expnsion around x = -5 yields:

    f(x) = -2/(x+5) + non-singular terms

    If we subtract the singular terms from
    f(x) we will be left with a rational functuon without any singularities, which is therefore a polynomial.

    Let's see what we get:

    f(x) - 2/(x+1) + 2/(x+5) =


    [x^4+6x^3+5x^2+8 - 2(x+5) + 2 (x+1)]/
    [(x+1)(x+5)] =

    [x^4 + 6 x^3 + 5 x^2]/[(x+1)(x+5)] =

    x^2 (x^2 + 6 x + 5)/[(x+1)(x+5)] =

    x^2

    So, we have:

    f(x) - 2/(x+1) + 2/(x+5) = x^2 ---->

    f(x) = x^2 + 2/(x+1) - 2/(x+5)

    The integral is thus:

    1/3 x^3 + 2 Log|x+1| - 2 Log|x+5| + c

  • Calculus -

    (x^4+6x^3+5x^2+8)/(x^2+6x+5)
    = x^2 + 8/(x^2 + 6x+5)
    = x^2 + 8/((x+1)(x+5))

    let 8/((x+1)(x+5) = A/(x+5) + B/(x+1)

    8 = A(x+1) + B(x+5)
    let x = -1
    8 = 4B --> B = 2
    let x = -5
    8 = -4A
    A = -2

    so ∫(x^4+6x^3+5x^2+8)/(x^2+6x+5) dx
    = ∫ (x^2 - 2/(x+5) + 2(x+1) dx
    = (1/3)x^3 - 2ln(x+5) + 2ln(x+1) + c

  • Calculus -

    Did not realize that Count Iblis had already answered this.
    I guess it helps to refresh the page before posting.

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