Thread: Simple integral is being reversed. Why?

When I integrate this:




I get:




However, when I plug this into Wolfram Alpha, it spits out this:




I have no idea why this is happening. A footnote in Wolfram Alpha says it's equivalent, but when I use the answer I get with the rest of the problem, I get an erroneous result - whereas the reversed version works fine.

I thought I understood how to take an integral of a simple fraction like this, but clearly I don't.

What am I missing?

Originally Posted by Lancet

When I integrate this:




I get:




However, when I plug this into Wolfram Alpha, it spits out this:




I have no idea why this is happening. A footnote in Wolfram Alpha says it's equivalent, but when I use the answer I get with the rest of the problem, I get an erroneous result - whereas the reversed version works fine.

I thought I understood how to take an integral of a simple fraction like this, but clearly I don't.

What am I missing?

absolute value ...

Answer to this:

What is integral of ?

This integration took place in an ODE problem as part of the integrating factor.



I was under the impression that the absolute value bars could be dropped in that circumstance...


Edit: And when I keep the absolute value bars in, it doesn't change the end result in any way. So that does not appear to be the root issue.

Originally Posted by Lancet

This integration took place in an ODE problem as part of the integrating factor.



I was under the impression that the absolute value bars could be dropped in that circumstance...


Edit: And when I keep the absolute value bars in, it doesn't change the end result in any way. So that does not appear to be the root issue.

f(x):R-->R

f(x)=ln(x) defined only when x>0.

Originally Posted by Also sprach Zarathustra

f(x):R-->R

f(x)=ln(x) defined only when x>0.

I'm afraid I don't understand. Or rather, I understand that ln(x) is defined only for positive numbers, but I don't understand how that explains why the terms appear to need to be reversed.

As a solution to the given integral, neither -3ln(60- t)+ C nor -3ln(t- 60)+ C is correct. What you should have is -3 ln|60-t|+ C or -3ln|t- 60|+ C. Those are equal, of course.

Now, when you looking for an integrating factor, you can set "C" equal to 0 and take the exponential: and you still need the absolute value. Since you are looking for a single integrating factor out of the infinite possible integrating factors (which is why we can set C equal to 0), you can use either or . Either will work because and you will be multiplying both sides of the equation by that so the negatives would cancel.