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Integration by substitution help

Integrate 2x/(2x-1)^2 using u=2x-1


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Reply 1
Avatar for TenOfThem
TenOfThem
18
Original post by SDavis123
Integrate 2x/(2x-1)^2 using u=2x-1


And you have tried ... ... ...
Reply 2
Avatar for SDavis123
SDavis123
OP
15
Original post by TenOfThem
And you have tried ... ... ...


This is what I did:


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(edited 11 years ago)
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Reply 3
Avatar for SDavis123
SDavis123
OP
15
I suppose its clearer to see if I show the step I skipped

(Integral) x/(2x-1)^2 . 2dx

Then substitute


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Reply 4
Avatar for TenOfThem
TenOfThem
18
Original post by SDavis123
This is what I did:


Fine until

u+12u2\dfrac{u+1}{2u^2}

Then you did nonsense
(edited 11 years ago)
Reply 5
Avatar for SDavis123
SDavis123
OP
15
Original post by TenOfThem
Fine until

u+1u2\dfrac{u+1}{u^2}

Then you did nonsense


But isn't it (u+1)/2u^2? Or is that the same as what you did brcause the 2 cancels with the du as du=2dx


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Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by SDavis123
But isn't it (u+1)/2u^2? Or is that the same as what you did brcause the 2 cancels with the du as du=2dx


yep

but then you still did nonsense :smile:
Reply 7
Avatar for SDavis123
SDavis123
OP
15
Original post by TenOfThem
yep

but then you still did nonsense :smile:


but what's wrong with the canceling part?


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Reply 8
Avatar for TenOfThem
TenOfThem
18
Original post by SDavis123
but what's wrong with the canceling part?


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Seriously?

Check this out

34=2+12×2=12\dfrac{3}{4} = \dfrac{2+1}{2 \times 2} = \dfrac{1}{2}

That is what you did
Reply 9
Avatar for SDavis123
SDavis123
OP
15
Original post by TenOfThem
Seriously?

Check this out

34=2+12×2=12\dfrac{3}{4} = \dfrac{2+1}{2 \times 2} = \dfrac{1}{2}

That is what you did


Argh I have no idea what to do :frown:


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Original post by SDavis123
Argh I have no idea what to do :frown:



u+12u2=u2u2+12u2\dfrac{u+1}{2u^2} = \dfrac{u}{2u^2} + \dfrac{1}{2u^2}

:biggrin:
Reply 11
Avatar for SDavis123
SDavis123
OP
15
Original post by TenOfThem
u+12u2=u2u2+12u2\dfrac{u+1}{2u^2} = \dfrac{u}{2u^2} + \dfrac{1}{2u^2}

:biggrin:


Wow I feel thick :frown:

Thanks for your help though :smile:


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Original post by SDavis123
Wow I feel thick :frown:

Thanks for your help though :smile:



np

sadly it is an error that I see time and time again :frown:
Reply 13
Avatar for SDavis123
SDavis123
OP
15
Original post by TenOfThem
np

sadly it is an error that I see time and time again :frown:


Just another question, am I correct up to here? I'm unsure if I might have made the same mistake


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you have multiplied by 2x rather than dividing by 2x when substituting for dx
Reply 15
Avatar for SDavis123
SDavis123
OP
15
Original post by TenOfThem
you have multiplied by 2x rather than dividing by 2x when substituting for dx


I'm not sure, my thinking was:

If I multiply by 2x I obtain 2x^4
I know du=2xdx
Therefore all I have to do is find x^3 in terms of u which would be (u-3/4)^3/2

Then I obtain an integral in terms of u:
(Integral) ((u-3/4)^3/2)/u^1/2 du


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Original post by SDavis123
I'm not sure, my thinking was:

If I multiply by 2x I obtain 2x^4
I know du=2xdx
Therefore all I have to do is find x^3 in terms of u which would be (u-3/4)^3/2

Then I obtain an integral in terms of u:
(Integral) ((u-3/4)^3/2)/u^1/2 du


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I do not really understand any of that

dx=du2xdx = \dfrac{du}{2x}
Reply 17
Avatar for Phichi
Phichi
11
Original post by SDavis123
I'm not sure, my thinking was:

If I multiply by 2x I obtain 2x^4
I know du=2xdx
Therefore all I have to do is find x^3 in terms of u which would be (u-3/4)^3/2

Then I obtain an integral in terms of u:
(Integral) ((u-3/4)^3/2)/u^1/2 du


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dx=du/2x. You cant decide to multiply by 2x :wink: Thus its x^3/2x say. Which is 1/2(x^2). So take 1/2 out of the integral, leaving you with 12∫U−4U\dfrac {1} {2}\int \dfrac {U-4} {\sqrt {U}}

Thank god
(edited 11 years ago)
Reply 18
Avatar for SDavis123
SDavis123
OP
15
This is a pretty similar question, except here I made dx the subject and just substituted, is that ok to do?



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Original post by SDavis123
This is a pretty similar question, except here I made dx the subject and just substituted, is that ok to do?



TBH I do not know what you were doing previously

You want dx as the subject as that is what you are trying to replace

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