Draper7824
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Need help with this question:

Integrate dx/sqrt(sqrt(x)+1) using the substitution u = sqrt(x)+1

I've been give the answer of 4/3(sqrt(x)-2)sqrt(sqrt(x)+1)+C, just need help getting there

Thanks guys!
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TeeEm
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(Original post by Draper7824)
Need help with this question:

Integrate dx/sqrt(sqrt(x)+1) using the substitution u = sqrt(x)+1

I've been give the answer of 4/3(sqrt(x)-2)sqrt(sqrt(x)+1)+C, just need help getting there

Thanks guys!
have seen substitution before?
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Draper7824
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(Original post by TeeEm)
have seen substitution before?
Yeh..
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User1443542
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(Original post by Draper7824)
Yeh..
Then I'm struggling to see where the difficulty is: Integrate u^.5 du/dx dx
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rayquaza17
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(Original post by Draper7824)
Need help with this question:

Integrate dx/sqrt(sqrt(x)+1) using the substitution u = sqrt(x)+1

I've been give the answer of 4/3(sqrt(x)-2)sqrt(sqrt(x)+1)+C, just need help getting there

Thanks guys!
You said you've seen substitutions before, it's just the same method.

u=sqrt(x)+1

Work out du/dx and then do the substitution that you're used to.

Note that when you do du/dx, you can use your substitution u=sqrt(x)+1 in that [might need to rearrange first though!]

Please post your working so far if you are still stuck.
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TeeEm
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(Original post by Draper7824)
Yeh..
then post your workings and I will try to see where you are going wrong
(I did get the answer that you claim without much difficulty)
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Draper7824
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Could be because it's Friday and I really can't be bothered but anyway..

found du/dx to be 1/(2u), so dx = 2u du (maybe ahahaha) then did some integration stuff and ended up with 2* the integral of u/sqrt(u) du which is probably totally wrong and now I've hit a brick wall and can't think about it
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poorform
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(Original post by Draper7824)
Could be because it's Friday and I really can't be bothered but anyway..

found du/dx to be 1/(2u), so dx = 2u du (maybe ahahaha) then did some integration stuff and ended up with 2* the integral of u/sqrt(u) du which is probably totally wrong and now I've hit a brick wall and can't think about it
just post your working in latex. nobody can really help you unless you do.
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Draper7824
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This is what I did which I know is totally wrong so you don't have to point that out
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rayquaza17
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(Original post by Draper7824)
.
It says in the question u=sqrt(x)+1, but you've used u=sqrt(x+1)?
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Draper7824
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Seems like I may have gone full retard #awkward I blame it on the fact it's the last day of teaching today and I'm going home tomorrow..


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Draper7824
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Can someone post their worked solution please, I still can't quite get it and can't really be bothered thinking about where I have gone wrong


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ztibor
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(Original post by Draper7824)
Can someone post their worked solution please, I still can't quite get it and can't really be bothered thinking about where I have gone wrong


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\displaystyle \int \frac{dx}{\sqrt{\sqrt{x}+1}}

with the given substitution you will get generally dx=f(u)*du and substituting it in dx you vill get only u functions in integrandus (but dx=g(x)*du form cause some mismatch in the variables ).

So
\displaystyle u=\sqrt{x}+1 from this \displaystyle (u-1)^2=x

and

\displaystyle \frac{dx}{du}=2\cdot(u-1) \rightarrow dx=2\cdot(u-1)\cdot du

\displaystyle 2\int \frac{u-1}{\sqrt{u} }du =2\int \left (u^{\frac{1}{2}}-u^{-\frac{1}{2}}\right )du

Use reverse differentiation
Generally
\displaystyle \int u^r du=\frac{u^{r+1}}{r+1}+C
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