Which of the following integrals cannot be integrated using partial fractions using linear factors with real coefficients?
a) integral of (x^2-1)/(x^3+x) dx
b) integral of 1/(9x^2-4) dx
c) integral of (x^3-x+3)/(x^2+x-2) dx
d) All of these can be integrated using partials fractions with linear factors and real coefficients
oops. I didn't see the part where it wanted only linear factors. So, the first one fails the test.
clearly they can all be handled using partial fractions.
All of the denominators can be broken in linear and quadratic factors.
(x^3+x) = x(x^2+1)
(9x^2-4) = (3x-2)(3x+2)
(x^2+x-2) = (x+2)(x-1)
It works if for the second fraction you use a linear expression, that is
(x^2-1)/(x^3+x) = (x^2-1)/((x)(x^2+1) )
let (x^2-1)/((x)(x^2+1) ) = A/x + (Bx+C)/(x^2 + 1)
A(x^2 + 1) + x(Bx + C) = x^2 - 1
let x = 0 ---> A + 0 = -1 or A = -1
let x = 1 -----> 2A + B+C = 0
B+C = 2 **
let x = -1 ---> 2A - (-B+C) = 0
B-C = 2 ***
add ** and *** ----> 2B = 4
B = 2 and C = 0
so (x^2-1)/(x^3+x) = -1/x + (2x)/(x^2 + 1)
∫(x^2-1)/(x^3+x) dx
= ∫ 2x/(x^2 + 1) dx - ∫ 1/x dx
= ln(x^2 + 1) - lnx + c
To determine which of the given integrals cannot be integrated using partial fractions with linear factors and real coefficients, we need to check if there are any irreducible quadratic factors in the denominators.
a) integral of (x^2-1)/(x^3+x) dx:
The denominator x^3+x does not have any irreducible quadratic factors, so partial fractions can be used.
b) integral of 1/(9x^2-4) dx:
The denominator 9x^2-4 can be factored as (3x+2)(3x-2), which contains irreducible quadratic factors. Therefore, partial fractions with linear factors and real coefficients cannot be used.
c) integral of (x^3-x+3)/(x^2+x-2) dx:
The denominator x^2+x-2 can be factored as (x+2)(x-1), which contains linear factors. However, it does not have any irreducible quadratic factors. Thus, partial fractions with linear factors and real coefficients can be used.
Therefore, the correct answer is:
b) integral of 1/(9x^2-4) dx