+270 I would greatly appreciate some guidance on how to go about solving for the following integral using the method of partial fractions..
\(\int \frac{dx}{x^3-x}\)
Take the integral:
integral1/(x^3 - x) dx
For the integrand 1/(x^3 - x), use partial fractions:
= integral(-1/x + 1/(2 (x + 1)) + 1/(2 (x - 1))) dx
Integrate the sum term by term and factor out constants:
= 1/2 integral1/(x + 1) dx - integral1/x dx + 1/2 integral1/(x - 1) dx
For the integrand 1/(x + 1), substitute u = x + 1 and du = dx:
= 1/2 integral1/u du - integral1/x dx + 1/2 integral1/(x - 1) dx
The integral of 1/u is log(u):
= (log(u))/2 - integral1/x dx + 1/2 integral1/(x - 1) dx
The integral of 1/x is log(x):
= -log(x) + (log(u))/2 + 1/2 integral1/(x - 1) dx
For the integrand 1/(x - 1), substitute s = x - 1 and ds = dx:
= -log(x) + (log(u))/2 + 1/2 integral1/s ds
The integral of 1/s is log(s):
= (log(s))/2 + (log(u))/2 - log(x) + constant
Substitute back for s = x - 1:
= (log(u))/2 + 1/2 log(x - 1) - log(x) + constant
Substitute back for u = x + 1:
= 1/2 log(x - 1) - log(x) + 1/2 log(x + 1) + constant
Factor the answer a different way:
= 1/2 (log(x - 1) - 2 log(x) + log(x + 1)) + constant
An alternative form of the integral is:
= 1/2 (log(x^2 - 1) - 2 log(x)) + constant
Which is equivalent for restricted x values to:
Answer: |= 1/2 log(1 - x^2) - log(x) + constant
