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