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Integration by substitution question.

The question is as follows;

Integrate ∫xx−2 dx\int \frac{x}{\sqrt {x-2}}\ dx , using the substitution x−2\sqrt {x-2}

I can't for the life of me figure out how to do this by substitution. Had it asked to integrate by parts I think it would be much easier.

I've had a go at it, but this is as far as I got.

dx=du(x−2)dx=du(\sqrt {x-2})

So after replacing dx I got

I= ∫2u2+4 dx\int 2u^2+4\ dx

And after replacing u, I got

23(x−2)32+4(x−2)12+c\frac{2}{3}(x-2)^\frac{3}{2}+4(x-2)^\frac{1}{2}+c

Which is really wrong. I'd appreciate it if anyone could point out where I went wrong. Thanks! :colondollar:
(edited 10 years ago)
Reply 1
Avatar for usycool1
usycool1
19
Original post by lebron_23

dx=du(x−2)dx=du(\sqrt {x-2})


How did you get this? :smile:

EDIT: Actually, looks like it's fine after I've just done it... :tongue:
(edited 10 years ago)
Reply 2
Avatar for atsruser
atsruser
11
Original post by lebron_23
The question is as follows;

Integrate ∫xx−2 dx\int \frac{x}{\sqrt {x-2}}\ dx , using the substitution u−2\sqrt {u-2}


I guess you mean u=x−2u =\sqrt{x-2}

1. If u=x−2u =\sqrt{x-2}, then x=f(u)x = f(u). What is f(u)f(u)?
2. If x=f(u)x = f(u), then what is dxdu\frac{dx}{du}, and hence what is dxdx?
Reply 3
Avatar for lebron_23
lebron_23
OP
2
Original post by usycool1
How did you get this? :smile:


Well if u=x−2u=\sqrt {x-2} then surely dudx=12x−2\frac{du}{dx}=\frac{1}{2\sqrt {x-2}}

Then I rearranged to get what dx was. Did I make an error differentiating? :facepalm:

Original post by atsruser
I guess you mean u=x−2u =\sqrt{x-2}

1. If u=x−2u =\sqrt{x-2}, then x=f(u)x = f(u). What is f(u)f(u)?
2. If x=f(u)x = f(u), then what is dxdu\frac{dx}{du}, and hence what is dxdx?


Ah yes you're quite right, I put a u instead of x. :eek:

And I'm not sure, I showed what I thought dx was but maybe that's wrong?
(edited 10 years ago)
Reply 4
Avatar for lebron_23
lebron_23
OP
2
Original post by Felix Felicis
I think you've just carried out the substitution incorrectly. What was your substitution? Was it x=u−2x = \sqrt{u - 2}? That won't help you very much, it'll transform your integral to something quite nasty. Try u=x−2u = \sqrt{x - 2}.


Aye, I wrote the substitution wrong in the OP, but I used the substitution u=x−2u = \sqrt{x - 2} in my working.
Reply 5
Avatar for usycool1
usycool1
19
Original post by lebron_23
Well if u=x−2u=\sqrt {x-2} then surely dudx=12x−2\frac{du}{dx}=\frac{1}{2\sqrt {x-2}}

Then I rearranged to get what dx was. Did I make an error differentiating? :facepalm:


Your differentiation is correct, but you rearranged incorrectly. :smile:
Reply 6
Avatar for atsruser
atsruser
11
Original post by Dalek1099
You look to have done it correctly.


Yes, in fact, the OP seems to have the correct answer anyway, though the layout is odd.
Reply 7
Avatar for lebron_23
lebron_23
OP
2
Original post by Dalek1099
You look to have done it correctly.


I sense a moment of stupidy on my behalf..

But the answer says 23(x+4)x−2+c\frac{2}{3}(x+4)\sqrt {x-2}+c. Is that the same as what I've got? It looks a world apart..

Original post by usycool1
Your differentiation is correct, but you rearranged incorrectly. :smile:


Oh I see. Perhaps I should go back to the drawing board lol
(edited 10 years ago)
Original post by lebron_23
Aye, I wrote the substitution wrong in the OP, but I used the substitution u=x−2u = \sqrt{x - 2} in my working.

Actually, upon doing the integral myself, you've got the correct answer, just need to simplify it a bit.
(edited 10 years ago)
Reply 9
Avatar for aveytare
aveytare
I'm pretty sure you're not wrong. http://www.wolframalpha.com/input/?i=integrate+x%2Fsqrt%28x-2%29
Reply 10
Avatar for lebron_23
lebron_23
OP
2
Original post by Felix Felicis
Actually, upon doing the integral myself, you've got the correct answer, just need to simplify it a bit.


Oh yes, I see it now.

Thank you very much for your help guys!

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