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?
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
3 answers
so will be answer C????
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."
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."
3 answers