Calculus

Which of the following integrals can be integrated using partial fractions using linear factors with real coefficients?
a) integral 1/(x^4-1) dx
b) integral (3x+1)/(x^2+6x+8) dx
c) integral x^2/(x^2+4)
d) None of these

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asked by Alice
  1. 1/(x^4-1) = a/(x^2+1) + b/(x-1) + c(x+1) <----- we can do that

    (3x+1)/(x^2+6x+8)
    = (3x+1)/((x+2)(x+4)) <--- yup, that will work

    x^2/(x^2+4) , mmmhhh, the bottom does not factor, and top and bottom have the same degree
    we could do x^2/(x^2+4) = 1 - 4/(x^2 + 4) , which is no better
    probably needs trig substitution

    so, what is your conclusion?
    Do you actually have to split them, or just find out if possible?

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  2. so will be answer C????

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    posted by Alice
  3. by linear factors, do you mean terms whose denominators are linear expressions such as

    A/(x-a)

    ?

    If so, then these can only work if the denominators do not contain irreducible quadratic factors such as ax^2+bx+c where the discriminant is negative. In such cases, the partial fraction terms are of the form
    (Ax+B)/(ax^2+bx+c)
    Now, in that case, the numerators are linear expressions.

    Not sure just what you mean by partial fractions with "linear factors."

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