Finding Integral of a Divided Function (Hyperbola?)

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The discussion focuses on integrating the function ∫((y+0.5)/(0.5-y)) dy between y=0 and y=-0.5. Participants suggest using polynomial long division to simplify the integration process, noting that the numerator and denominator are of the same degree. One approach involves rewriting the function to separate it into a quotient and a remainder, allowing for easier integration. There is confusion about how to integrate with a remainder, but clarification is provided on expressing the remainder as a fraction. Overall, the conversation emphasizes the importance of simplifying the function for successful integration.
Mathpower
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Homework Statement


Integrate between y=0 and y=-0.5:
∫((y+0.5)/(0.5-y)) dy

Homework Equations


Can you please show me how to integrate it...Then I will be able to take it from there and substitute in the appropriate values.


The Attempt at a Solution


Quotient rule in reverse?...wouldn't work
Using ln ...wouldn't work
Try to factorise...nope
Use rationalising of denominator (conjugate idea)...that just makes it even more complicated
Now I have run out of ideas ...this is out of my scope. (Scholarship paper NZ)

Thank you in advance,
Kindest Regards.
 
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Mathpower said:

Homework Statement


Integrate between y=0 and y=-0.5:
∫((y+0.5)/(0.5-y)) dy

Homework Equations


Can you please show me how to integrate it...Then I will be able to take it from there and substitute in the appropriate values.


The Attempt at a Solution


Quotient rule in reverse?...wouldn't work
Using ln ...wouldn't work
Try to factorise...nope
Use rationalising of denominator (conjugate idea)...that just makes it even more complicated
Now I have run out of ideas ...this is out of my scope. (Scholarship paper NZ)

Thank you in advance,
Kindest Regards.

Hint: Divide the numerator by the denominator to make the degree of the numerator one less than the degree of the denominator.
 
Hmmm...Thanks for the hint, but I don't get what you meant here...
Isn't the numerator already being divided by the denominator and aren't they to the same degree anyway?
 
Write it as$$
-\frac{y + \frac 1 2}{y -\frac 1 2}$$
and do one step of a long division writing it as quotient + remainder/divisor.

Alternatively you could write it as$$
-\frac {(y-\frac 1 2)+1}{(y-\frac 1 2)}$$and break it into two terms. It's the same result either way.
 
Mathpower said:
Hmmm...Thanks for the hint, but I don't get what you meant here...
Isn't the numerator already being divided by the denominator and aren't they to the same degree anyway?
Yes, they're the same degree, so use some sort of division algorithm to find a quotient & remainder.
 
Nice! I like the alternative (because its the only thing I understood) from LCKurtz...rationalising the denominator.
The rest of it didn't make much sense...how can you integrate with a remainder! (I must be missing something)
Thank you so much for all your help.

P.S.: Sorry for not being able to understand what you guys said.

Kindest Regards
 
Mathpower said:
Nice! I like the alternative (because its the only thing I understood) from LCKurtz...rationalising the denominator.
The rest of it didn't make much sense...how can you integrate with a remainder! (I must be missing something)
Thank you so much for all your help.

P.S.: Sorry for not being able to understand what you guys said.

Kindest Regards
Here's how that remainder works:

Write the remainder of 1 as a fraction.

\displaystyle \frac{1}{y-\frac{1}{2}}

So that \displaystyle -\frac {(y+\frac 1 2)}{(y-\frac 1 2)}=-\left(1+\frac{1}{y-\frac{1}{2}}\right)

Although expressions like the one here, I often use the same method as LCKurtz showed you. For more complicated situations, you should know how to express the remainder as a fraction.
 

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