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