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

Answers

Answered by oobleck
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)
Answered by oobleck
oops. I didn't see the part where it wanted only linear factors. So, the first one fails the test.
Answered by Reiny
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
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