# Thread: Simple integral is being reversed. Why?

1. ## Simple integral is being reversed. Why?

When I integrate this:

$\displaystyle \int \frac{3}{60 - t} dt$

I get:

$\displaystyle -3ln(60 - t) + c$

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

$\displaystyle -3ln(t - 60) + c$

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?

2. ## Re: Simple integral is being reversed. Why?

Originally Posted by Lancet
When I integrate this:

$\displaystyle \int \frac{3}{60 - t} dt$

I get:

$\displaystyle -3ln(60 - t) + c$

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

$\displaystyle -3ln(t - 60) + c$

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 ...

$\displaystyle \int \frac{3}{60-t} \, dt = -3\ln|60-t| + C = -3\ln|t-60| + C$

3. ## Re: Simple integral is being reversed. Why?

Answer to this:

What is integral of $\displaystyle \frac{1}{-x}$ ?

4. ## Re: Simple integral is being reversed. Why?

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

$\displaystyle U = e^{\int \frac{3}{60 - t} dt} = (60 - t)^{-3}$

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.

5. ## Re: Simple integral is being reversed. Why?

Originally Posted by Lancet
This integration took place in an ODE problem as part of the integrating factor.

$\displaystyle U = e^{\int \frac{3}{60 - t} dt} = (60 - t)^{-3}$

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.

6. ## Re: Simple integral is being reversed. Why?

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.

7. ## Re: Simple integral is being reversed. Why?

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: $\displaystyle e^{-3ln|60-t|}= |60- t|^{-3}$ 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 $\displaystyle (60- t)^{-3}$ or $\displaystyle (t- 60)^{-3}$. Either will work because $\displaystyle (60- t)^{-3}= -(t- 60)^{-3}$ and you will be multiplying both sides of the equation by that so the negatives would cancel.