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I thought I could use substitution but I can't ...
is this right? \[\int\limits_{0}^{4}\frac{ x }{ \sqrt{1+2x} }\]

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Other answers:

yep :) .
put u= 1 + 2 x
\( u = 1+ 2x => du/dx = 2\) \( 1+2x = u => 2x = u-1 => x = (u-1)/2 \)
I got to that step. Then I got stuck :P .
Maybe:\[\int\limits\frac{x}{\sqrt{1+2x}} dx => \int\limits \frac{(u-1)/2}{\sqrt{u}} *2du \int\limits\frac{u-1}{2} *\frac{1}{\sqrt{u}} 2du => 2\int\limits\frac{u-1}{2} *\frac{1}{\sqrt{u}} \]
Here what wolfram alpha gave
1 Attachment
Would my step work tho?
@Dido525 do you understand what mimi did? those are the correct steps to take.
Yeah I actually do. I just need to process it.
@Mimi_x3 : Wouldn't you write du/2 instead of 2*du ?
Because we said du/2 was dx.
Oh yes :o good call.
Yeah typo lol sorry! that was where i went wrong :(
It's okay :) .
So so far I have: \[\frac{ 1 }{ 4 }\int\limits_{1}^{9} \frac{ u-1 }{ \sqrt{u} } du\]
then break it up!
Ohh... Right :P .
integrate it; then don't forget to sub back in \(u\)
Thanks! :) .

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