Long Division for Rational Expression Integration

[Jason76]

Unless of course, you have a rational expression with only \(\displaystyle 1\) in the numerator, or if you have an \(\displaystyle x\) term and a constant (in the numerator), OR the \(\displaystyle x\) term already matches the \(\displaystyle du\) in the denominator, OR the terms in the numerator match the du (or some other easily manipulated problem).

Even long division may be difficult, it might be a more "straightforward" and "easy to understand" way of doing problems like that below:

http://www.freemathhelp.com/forum/threads/81280-Integration-Problem

 

[mmm4444bot]

I'm not understanding the post above. Are you working on a particular exercise?

If so, please state the problem and explain the meaning of du in the numerator and du in the denominator. I'm not sure what that means, either.

Thank you.

 

[Jason76]

I'm not understanding the post above. Are you working on a particular exercise?

If so, please state the problem and explain the meaning of du in the numerator and du in the denominator. I'm not sure what that means, either.

Thank you.

Actually, I did some research, and the "long division method of integration" only works when the highest index in the numerator is greater or equal to the highest index in the denominator.

What I mean by \(\displaystyle dx = du\) is this: \(\displaystyle \int \dfrac{2x - 2}{x^{2} - 2x} dx\) would be the case of \(\displaystyle dx = du\). Also in the case of, \(\displaystyle \int \dfrac{dx}{some-expression}\), you would just use some of the memorized integral formulas.

Not working on a particular exercise, but just wanted some opinions on whether this was a good route to take. But in the case of it NOT being "numerator is greater or equal to the highest index in the denominator", then another method is applicable. Maybe partial fractions? But either way, based on the fraction integration threads I submitted, I don't want to use those methods (that were used in the threads) to solve it, if another way is there.

Let's look at

\(\displaystyle \int \dfrac{5x - 2}{\sqrt{x^{2} + 2x}} dx \)

In order to use the long division method, first we need to get rid of the square root, and later see if the index numbers fit with "long division". But you could even go another route and de-fraction the problem by getting rid of the square root (by expressing the \(\displaystyle 1/2\) exponent as negative). In that case, you wouldn't worry with long division anyways.

 

[Deleted member 4993]

Actually, I did some research, and the "long division method of integration" only works when the highest index in the numerator is greater or equal to the highest index in the denominator.

What I mean by \(\displaystyle dx = du\) is this: \(\displaystyle \int \dfrac{2x - 2}{x^{2} - 2x} dx\) would be the case of \(\displaystyle dx = du\). Not working on a particular exercise, but just wanted some opinions on whether this was a good route to take. But in the case of it NOT being "numerator is greater or equal to the highest index in the denominator", then another method is applicable. Maybe partial fractions? But either way, based on the fraction integration threads I submitted, I don't want to use those methods (that were used in the threads) to solve it, if another way is there.

Let's look at

\(\displaystyle \int \dfrac{5x - 2}{\sqrt{x^{2} + 2x}} dx \)

In order to use the long division method, first we need to get rid of the square root, and later see if the index numbers fit with "long division".

In my opinion the best route to calculate:


\(\displaystyle \int \dfrac{5x - 2}{\sqrt{x^{2} + 2x}} dx \)

is to complete square in the denominator:

\(\displaystyle \int \dfrac{5(x+1) - 7}{\sqrt{(x+1)^{2} - 1}} dx \)

and execute trigonometric substitution:

x + 1 = sec(Θ)

and continue...

 

[Jason76]

In my opinion the best route to calculate:


\(\displaystyle \int \dfrac{5x - 2}{\sqrt{x^{2} + 2x}} dx \)

is to complete square in the denominator:

\(\displaystyle \int \dfrac{5(x+1) - 7}{\sqrt{(x+1)^{2} - 1}} dx \)

and execute trigonometric substitution:

x + 1 = sec(Θ)

and continue...

In any case, your method is better than gasping at an answer using my book's method of "hammering \(\displaystyle du\) into \(\displaystyle dx\)".