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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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

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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