# Integrate dx/sqrt(sqrtx+1)Watch

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#1
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!
0
4 years ago
#2
(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?
0
#3
(Original post by TeeEm)
have seen substitution before?
Yeh..
0
4 years ago
#4
(Original post by Draper7824)
Yeh..
Then I'm struggling to see where the difficulty is: Integrate u^.5 du/dx dx
0
4 years ago
#5
(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!]

0
4 years ago
#6
(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)
0
#7
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
0
4 years ago
#8
(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
0
#9
This is what I did which I know is totally wrong so you don't have to point that out
0
4 years ago
#10
(Original post by Draper7824)
.
It says in the question u=sqrt(x)+1, but you've used u=sqrt(x+1)?
0
#11
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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0
#12
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
0
4 years ago
#13
(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

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
1
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