Find the integral of x/(x-6) with substitution

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  • #1
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[tex]\int\frac{x}{x-6}dx[/tex]


[tex]u=x-6[/tex]

[tex]\int \frac{u+6}{u}du[/tex]

[tex]\int 1+\frac{6}{u} du[/tex]

[tex]u+6ln|u|+C[/tex]

[tex]x-6+6ln|x-6|+C[/tex]

this appears to be incorrect although it seems logical, also if somone could please tell me the syntax for the definte integral

also similar case here

[tex]\int \frac{x^2}{x+4}dx[/tex]

[tex]u=x+4[/tex]
[tex]\int \frac{(u-4)^2}{u}du[/tex]
[tex]\int \frac{u^2-8u+16}{u}du[/tex]
[tex]\int u-8+\frac{16}{u} du[/tex]
[tex]\frac{u^2}{2}-8u+16ln|u|+C[/tex]
[tex]\frac{(x+4)^2}{2}-8(x+4)+16ln|x+4|+C[/tex]
 
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Answers and Replies

  • #2
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When do you a u-sub don't forget to find du.

Hint on why that sub won't work. d/du of x-6 isn't x.
 
  • #3
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in the first problem

x=u+6 and du=dx
 
  • #4
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For the second problem, have you considered Polynomial long division, or do you have to use a substitution?
 
  • #5
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in the first problem

x=u+6 and du=dx

How'd you end up with the u+6 on the top?

Ohh...I see what you did.

The only thing I can think of where the mistake was made. I tried it using long division first and got the correct answer.

Someone smarter than I will have to answer that as I don't solve integrals that way.
 
  • #6
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For the first, I think more algebra is required than substitutions, you just need to be cleaver.
 
  • #7
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If you are getting the same error in both of those, maybe your substitution trick is the problem.

Is that even allowable?
 
  • #8
Char. Limit
Gold Member
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Actually, it is allowable... look up integration by substitution on wikipedia.

Your first answer is right. Just remember that since C is any constant number, the -6 is merged into C to give...

[tex]x+6log(x-6)+C[/tex]
 
  • #9
34,904
6,647


[tex]\int\frac{x}{x-6}dx[/tex]


[tex]u=x-6[/tex]

[tex]\int \frac{u+6}{u}du[/tex]

[tex]\int 1+\frac{6}{u} du[/tex]

[tex]u+6ln|u|+C[/tex]

[tex]x-6+6ln|x-6|+C[/tex]

this appears to be incorrect although it seems logical, also if somone could please tell me the syntax for the definte integral
The first one can be done by polynomial long division or by substitution. I don't see anything wrong with your answer. If you think that your answer is incorrect, it might only be that your answer differs from the "correct" answer by a constant. If you take the derivative of your result, you get the original integrand. erok81 is not giving you good advice.

For the second problem, one approach is to use polynomial long division, although your substitution works just fine. Again, the two different approaches give differnent appearing antiderivatives, but they differ only by a constant.
also similar case here

[tex]\int \frac{x^2}{x+4}dx[/tex]

[tex]u=x+4[/tex]
[tex]\int \frac{(u-4)^2}{u}du[/tex]
[tex]\int \frac{u^2-8u+16}{u}du[/tex]
[tex]\int u-8+\frac{16}{u} du[/tex]
[tex]\frac{u^2}{2}-8u+16ln|u|+C[/tex]
[tex]\frac{(x+4)^2}{2}-8(x+4)+16ln|x+4|+C[/tex]
 
  • #10
34,904
6,647


When do you a u-sub don't forget to find du.
He didn't forget it; he just didn't explicitly show how he found du
Hint on why that sub won't work. d/du of x-6 isn't x.
And that's not what he did. Why would you want to take the derivative with respect to u of x - 6? That doesn't make any sense in this problem.
 
  • #11
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The first one can be done by polynomial long division or by substitution. I don't see anything wrong with your answer. If you think that your answer is incorrect, it might only be that your answer differs from the "correct" answer by a constant. If you take the derivative of your result, you get the original integrand. erok81 is not giving you good advice.

There is more than one way to solve this problem. He thought his method was wrong, so I did it my way - which wasn't a u-sub to start with. So it wasn't bad advice, just a different way to solve it. Notice I did say someone smarter than I, like you, needs to come in and clarify since I wouldn't solve this problem the way he did. :)

He didn't forget it; he just didn't explicitly show how he found du

And that's not what he did. Why would you want to take the derivative with respect to u of x - 6? That doesn't make any sense in this problem.

He did take the deritive of his u-sub and got dx. I commented on it and he explained.

To make the last part clear so I make mistakes on my homework. If you have a problem say x/x^2+6. Your u-sub is u=x^2+6 then you take the deritive and get du = 2x dx or du/2 = x dx. What is wrong with my method? Sure it skips 2-3 steps, but it isn't wrong. Is it? I hope not since I've made it through calc I and most of II doing them that way.
 
  • #12
34,904
6,647


There is more than one way to solve this problem. He thought his method was wrong, so I did it my way - which wasn't a u-sub to start with. So it wasn't bad advice, just a different way to solve it. Notice I did say someone smarter than I, like you, needs to come in and clarify since I wouldn't solve this problem the way he did. :)



He did take the deritive of his u-sub and got dx. I commented on it and he explained.

To make the last part clear so I make mistakes on my homework. If you have a problem say x/x^2+6. Your u-sub is u=x^2+6 then you take the deritive and get du = 2x dx or du/2 = x dx. What is wrong with my method? Sure it skips 2-3 steps, but it isn't wrong. Is it? I hope not since I've made it through calc I and most of II doing them that way.
No, that's fine. The bad advice was telling the OP why a substitution won't work (it does work), and telling him/her that d/du (x - 6) isn't x. Maybe you meant d/dx(x - 6).
 

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