# Integrate dx/sqrt(sqrtx+1) Watch

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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!

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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#2

(Original post by

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!

**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!

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Yeh..

**Draper7824**)Yeh..

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#5

**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!

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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#6

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Yeh..

**Draper7824**)Yeh..

(I did get the answer that you claim without much difficulty)

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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

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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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)

**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)

__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__**then did some integration stuff**
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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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#10

(Original post by

.

**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..

Posted from TSR Mobile

Posted from TSR Mobile

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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

Posted from TSR Mobile

Posted from TSR Mobile

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#13

(Original post by

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

Posted from TSR Mobile

**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

Posted from TSR Mobile

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

from this

and

Use reverse differentiation

Generally

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