Integrating 1/(1+sqrt(2x)) using u-substitution

[johnsonandrew]

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...

 

[rocomath]

Well from rationalizing we get ...

$$\int\left(\frac{1}{1-2x}-\frac{\sqrt{2x}}{1-2x}}\right)dx$$

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

$$-\int\frac{\sqrt{2x}}{1-2x}dx$$

$$u=\sqrt{2x}\rightarrow u^2=2x$$

$$u^2=2x \leftrightarrow udu=dx$$

 

[johnsonandrew]

following you so far

 

[rocomath]

following you so far

After substituting, we get ...

$$\int\frac{-u^2}{1-u^2}du$$

Then add $$\pm 1$$ to the numerator so that you can split it into 2.

$$\int\frac{(1-u^2)-1}{1-u^2}du$$

 

[johnsonandrew]

I don't understand:
$$u^2=2x \leftrightarrow udu=dx$$

 

[rocomath]

I don't understand:
$$u^2=2x \leftrightarrow udu=dx$$

I made my initial u-sub then I manipulated my u-sub by squaring both sides and then I took it's derivative.

$$u=\sqrt 2x$$ ONLY for the numerator

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

$$u^2=2x$$

Taking the derivative of my manipulating u-sub

$$2udu=2dx \rightarrow udu=dx$$

 

[johnsonandrew]

Ohhh okay. Thank you!

 

[rocomath]

Ohhh okay. Thank you!

Anytime.

 

[rocomath]

Actually, rationalizing isn't even a good idea. You can apply the same methods I did with the u-sub w/o rationalizing.