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

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?

so will be answer C????

To determine which of the given integrals can be integrated using partial fractions with linear factors, we need to check if the denominator of each integral can be factored into linear factors with real coefficients.

a) integral 1/(x^4-1) dx:
To factorize x^4-1, we can use the difference of squares formula: a^2-b^2 = (a+b)(a-b).
Here, x^4-1 can be factored as (x^2+1)(x^2-1). However, x^2+1 has complex roots (x=±i), which means it cannot be factored into linear factors with real coefficients. Thus, partial fractions cannot be applied to this integral.

b) integral (3x+1)/(x^2+6x+8) dx:
We need to factorize the denominator (x^2+6x+8) to determine if it can be written as a product of linear factors. Factoring this quadratic expression, we get (x+4)(x+2). This is a valid factorization with linear factors and real coefficients, so partial fractions can be applied to this integral.

c) integral x^2/(x^2+4) dx:
In this case, the denominator x^2+4 cannot be factored into linear factors with real coefficients. It is in the form of a sum of squares, which does not allow for partial fractions to be applied. So, partial fractions cannot be done for this integral either.

d) None of these:
Based on the analysis above, the correct answer is (b) integral (3x+1)/(x^2+6x+8) dx since it is the only integral where the denominator can be factorized into linear factors with real coefficients, allowing us to apply partial fractions.

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."