Integrating Rational Expressions

[Jason76]

The quotient rule is used for the getting the derivatives of rational expressions, but with integration something different is done.

\(\displaystyle \int \dfrac{-2x}{e^{x} - 2} dx\)

Would we go this way?

\(\displaystyle \int \dfrac{-2x }{e^{x}(2) } - \dfrac{-2x}{2(e^{x})} dx\) - Logic being that the problem can be broken into two fraction each with a different denominator. But because of different denominators, there has to be an attempt to make the denominators alike.

\(\displaystyle \int \dfrac{-4x}{2e^{x}} - \dfrac{-2xe^{x}}{2e^{x}} dx\)

Right so far?




This book says to do something different.

 

[DrPhil]

The quotient rule is used for the getting the derivatives of rational expressions, but with integration something different is done.

\(\displaystyle \int \dfrac{-2x}{e^{x} - 2} dx\)

Would we go this way?

\(\displaystyle \int \dfrac{-2x }{e^{x}(2) } - \dfrac{-2x}{2(e^{x})} dx\) - Logic being that the problem can be broken into two fraction each with a different denominator. But because of different denominators, there has to be an attempt to make the denominators alike.

\(\displaystyle \int \dfrac{-4x}{2e^{x}} - \dfrac{-2xe^{x}}{2e^{x}} dx\)

Right so far? NO!

This book says to do something different.

Are your two fractions equal to the original integrand??
"Making denominators alike" would have to be (e^x - 2), and you can't get there fromn what you have.

Have you perhaps gone to a new chapter? When "this book says something different," might that include "integration by parts"?

 

[Deleted member 4993]

The quotient rule is used for the getting the derivatives of rational expressions, but with integration something different is done.

\(\displaystyle \int \dfrac{-2x}{e^{x} - 2} dx\)

Would we go this way?

No ..

To do the above problem - one of the ways would be to substitute:

u = e^x - 2 → x = ln(u+2) → dx = du/(u+2)

so your integrand becomes

\(\displaystyle 2\int \dfrac{ln(u+2)}{u}\, \dfrac{du}{u+2}\)

and continue....


\(\displaystyle \int \dfrac{-2x }{e^{x}(2) } - \dfrac{-2x}{2(e^{x})} dx\) - Logic being that the problem can be broken into two fraction each with a different denominator. But because of different denominators, there has to be an attempt to make the denominators alike.

\(\displaystyle \int \dfrac{-4x}{2e^{x}} - \dfrac{-2xe^{x}}{2e^{x}} dx\)

Right so far?




This book says to do something different.

,

 

[Jason76]

Are your two fractions equal to the original integrand??
"Making denominators alike" would have to be (e^x - 2), and you can't get there fromn what you have.

Have you perhaps gone to a new chapter? When "this book says something different," might that include "integration by parts"?

I can't understand the strategy used by Khan (poster on the thread) yet. Also, how can u do integration by parts on a quotient?

 

[DrPhil]

I can't understand the strategy used by Khan (poster on the thread) yet. Also, how can u do integration by parts on a quotient?

For integration by parts, you need to choose u and dv such that the integral looks like \(\displaystyle \int{u\ dv}\), hopefully with dv being something you can integrate and u being something you can differentiate. Since I know how to integrate \(\displaystyle dv = 2x\ dx\), that would leave \(\displaystyle u = -1/(e^x-2)\). That does not look particularly hopeful, because the derivative of u has a higher power in the denominator. Unless this is from a chapter on integration by parts, I fear that I may be leading you toward a wrong path.

I would then try the substitution Kahn suggested.
Let \(\displaystyle u = e^x - 2\)............\(\displaystyle du = e^x dx\)
.....\(\displaystyle x = \ln{(u + 2)}\).......\(\displaystyle dx = du/(u + 2)\)

Can you tell us what you did not understand about that substitution?

Please show us your work on either of these approaches! When you ask open-ended questions, it seems we are trying to teach the lesson instead of coaching you.