Integration of improper fraction using substitution

[Skelly4444]

Having trouble agreeing with the book on this one too.
Not quite sure I'm dealing with the quotient correctly.

I've attached my workings which hopefully will shed some light on it.

 

[BigBeachBanana]

You forgot the constant of integration. -2+C is another constant.

 

[topsquark]

Apparently you and the book did the problem in two different ways. Your answers are off by a constant, which is perfectly acceptable.

-Dan

 

[lex]

[imath]\dfrac{x(x-4)}{(x-2)^2} = \dfrac{x^2-4x}{(x-2)^2} =\dfrac{(x-2)^2-4}{(x-2)^2} = 1-\dfrac{4}{(x-2)^2}\\[/imath]
[math]\hspace{-30em} {\therefore \int \dfrac{x(x-4)}{(x-2)^2} \,dx = \int 1-\dfrac{4}{(x-2)^2} dx}\\ [/math][imath]\hspace27ex = x +\dfrac{4}{x-2} + c[/imath]

 

[lex]

I wasn't sure about the extra 2 that came from my calculation and whether or not we could just add it to the constant of integration and treat it as a single entity. Am I correct in assuming that both answers are correct then?

The thing that puzzles me is why would you go to the bother of using polynomial division to obtain the quotient and remainder when it can just be integrated as it stands using the substitution I've outlined?

If this reply refers to this thread,
yes, as the earlier posters say, both answers are correct and you can incorporate the -2 into the constant.
The book's method didn't use long division, but just simple algebra.
Your method is perfectly good and straightforward.

(In general in mathematics a solution is considered more 'elegant' if it uses simpler mathematics, rather than relying on a theorem or a more sophisticated method, as in 'integration by substitution'. The book's method might be considered slightly more 'beautiful' than using substitution.
However that is not important when doing exams. You want to use a method which gets you the answer accurately and efficiently!)