Integratind Same Simple Equation In Different Forms, Getting Different Results HELP

[RoyleWilliam]

Hello, my problem deals with the integration of the equation (2x-1)/(x(x-1))
The Equation (2x-1)/(x(x-1)) is Identical to the equation (1/(1-x)-1/x), as I have proved and confirmed adequately using partial fractions and through graphing.
However, when you integrate the equation, the answer seems to differ according to the form you integrate it in. Even using wolfram, I notice that the result differs, even though the equation is the same.

When I integrate (1/(1-x)-1/x) with respect to X, I end up with -ln(1-x)-ln(x)

Yet when I integrate (2x-1)/(x(x-1)) with respect to X, I end up with log(x-1)+log(x)

shouldn't I get the same result either way? I need help figuring this out. Thanks, itll get me out of my confusion

 

[Deleted member 4993]

Hello, my problem deals with the integration of the equation (2x-1)/(x(x-1))
The Equation (2x-1)/(x(x-1)) is Identical to the equation (1/(1-x)-1/x), as I have proved and confirmed adequately using partial fractions and through graphing.
However, when you integrate the equation, the answer seems to differ according to the form you integrate it in. Even using wolfram, I notice that the result differs, even though the equation is the same.

When I integrate (1/(1-x)-1/x) with respect to X, I end up with -ln(1-x)-ln(x)

Yet when I integrate (2x-1)/(x(x-1)) with respect to X, I end up with log(x-1)+log(x)

shouldn't I get the same result either way? I need help figuring this out. Thanks, itll get me out of my confusion

That is because:

(2x-1)/[x(x-1)] = 1/x + 1/(x-1)

The domain of the solutions are different

for your solution: 0<x<1 .....................-ln(1-x)-ln(x)

for the other solution x>1....................log(x-1)+log(x)

Log functions are funny - - just need to be careful about the domain of the solution.

 

[RoyleWilliam]

Thanks, another question that the earlier question started

Thanks for helping me figure that out. Your definately right, they are both right but partial solutions for different domain ranges. However, I'm still struggling with the problem that gave rise to this question. It's a first order differential equation, and I can't for the life of me find the solution. I can't find the factor of integration, in particular. Here is the problem

Solve The Equation
x(1-x^2)d[y]/d[x]+(2x^2-1)y=2x^3 for 0>x>1 (I understand what you mean about the relevance of domain for log functions)

That is the problem



What I've done so far:
d[y]/d[x]+(2x^2-1)/(x(1-x^2))y=2x^3/(x(1-x^2))
Then we must find the factor of integration, e to the power of the integral of
(2x^2-1)/(x(1-x^2)) with respect to X

What I do then is use the substitution u=x^2
then we end up taking the integral of
(2u-1)/(u(1-u))
with respect to u
using partial fractions, I convert it too the integral
1/(1-u)-1/u with respect to u

so the factor of integration should be e to the power of the integral of 1/(1-u)-1/u with respect to u
when I integrate with respect to u, I get C-ln(1-u)-ln(u) for 0>x>1. Because we are hunting for any avaliable factor of integration, we will suppose C=0
when we raise e to the power (after subsititing x^2 for u), we wind up with
1/(x*(1-x^2)^.5)

but that can't be the factor of integration, because the derivative of 1/(x*(1-x^2)^.5) is 1/(x^2*(1-xx^2)^.5), which is not the same as (2x^2-1)/(x(1-x^2))*1/(x*(1-x^2)^.5)

I know this is a pain, and thanks for helping me above, but it would be a lifesaver if you could tell me what the correct factor of integration is for this problem, and how you found it