Integral (1/(1+sqrt(2x))) dx

  • #1

Homework Statement


Find the indefinite integral.
∫ (1/(1+sqrt(2x))) dx

Homework Equations


∫ 1/u du = ln |u| + C


The Attempt at a Solution


I tried a couple 'u' substitutions, which didn't work out. I also tried rationalizing the denominator, but that didn't help. No one I've talked to knows how to do this one...
 

Answers and Replies

  • #2
1,752
1
Well from rationalizing we get ...

[tex]\int\left(\frac{1}{1-2x}-\frac{\sqrt{2x}}{1-2x}}\right)dx[/tex]

So from here, the left is easy and now we work only with the right

[tex]-\int\frac{\sqrt{2x}}{1-2x}dx[/tex]

[tex]u=\sqrt{2x}\rightarrow u^2=2x[/tex]

[tex]u^2=2x \leftrightarrow udu=dx[/tex]
 
Last edited:
  • #4
1,752
1
following you so far
After substituting, we get ...

[tex]\int\frac{-u^2}{1-u^2}du[/tex]

Then add [tex]\pm 1[/tex] to the numerator so that you can split it into 2.

[tex]\int\frac{(1-u^2)-1}{1-u^2}du[/tex]
 
Last edited:
  • #5
I don't understand:
[tex]u^2=2x \leftrightarrow udu=dx[/tex]
 
  • #6
1,752
1
I don't understand:
[tex]u^2=2x \leftrightarrow udu=dx[/tex]
I made my initial u-sub then I manipulated my u-sub by squaring both sides and then I took it's derivative.

[tex]u=\sqrt 2x[/tex] ONLY for the numerator

Manipulating my u-sub by squaring both sides so that I can substitute for my denominator.

[tex]u^2=2x[/tex]

Taking the derivative of my manipulating u-sub

[tex]2udu=2dx \rightarrow udu=dx[/tex]
 
  • #9
1,752
1
Actually, rationalizing isn't even a good idea. You can apply the same methods I did with the u-sub w/o rationalizing.
 

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