Maths work

[dttom]

The question is to find to integral of 1/(x(sqrt(x+4)))

I tried substitution with x equals 4(tan(x))^2 but not succeed, could anyone help?

[the tree]

Maybe you could start by working out [imath]\frac{\mbox{d}}{\mbox{d}x} \ln(\sqrt{x+a}+b)[/imath].

[ajb]

Tree has got the right idea. He knows from experience that the answer is something like what he suggests differentiating. It is not too hard to get the answer.

[the tree]

Oh ffs. Okay since you had to read that, whateverthehell, I'll be lax on the rules and give a complete answer.

 

You will of course need to fill in the gaps here and there.

 

This isn't an easy integral so bare with me...

 

Assuming you already know that [imath]\frac{\mbox{d}}{\mbox{d}x}\ln( f(x) )=\frac{f'(x)}{f(x)}[/imath] it should be easy enough to work out:

[math]\frac{\mbox{d}}{\mbox{d}x} \ln(\sqrt{x+4}+b)=\frac{1}{2\sqrt{x+4}}\cdot\frac{1}{\sqrt{x+4}+b}[/math].

 

Then you'll need to show that:

[math]\frac{1}{\sqrt{x+4}(\sqrt{x+4}+b_1)}-\frac{1}{\sqrt{x+4}(\sqrt{x+4}+b_2)}=\frac{b_2 - b_1}{\sqrt{x+4}(\sqrt{x+4}+b_1)(\sqrt{x+4}+b_2)}[/math]

which isn't anything more than basic high school algebra.

 

Thus far you can conclude:

[math]\frac{\mbox{d}}{\mbox{d}x} \left( \ln(\sqrt{x+4}+b_1) - \ln(\sqrt{x+4}+b_1) \right)=\frac{1}{2}\frac{b_2 - b_1}{\sqrt{x+4}(\sqrt{x+4}+b_1)(\sqrt{x+4}+b_2)}[/math]

 

Now you should see where I'm going, substitute in [imath]b_1=-2[/imath] and [imath]b_2=2[/imath].

 

[math]\frac{\mbox{d}}{\mbox{d}x} \left( \ln(\sqrt{x+4}-2) - \ln(\sqrt{x+4}+2) \right)=\frac{1}{2}\frac{4}{x\sqrt{x+4}}[/math]

 

And therefore, with just some scaling to finish up.

 

[math]\int \frac{1}{x\sqrt{x+4}} \mbox \,{d}x = \tfrac{1}{2}\left( \ln (\sqrt{x+4}-2 )-\ln ( \sqrt{x+4}+2 ) \right) + c[/math].

Edited May 4, 2010 by the tree

[ajb]

Nicely done the tree.

[Cap'n Refsmmat]

(Apologies for the slight loss in continuity here when we removed some gibberish. ajb's post up there makes a little less sense when the post it was responding to vanished...)

[Charlatan]

(Apologies for the slight loss in continuity here when we removed some gibberish. ajb's post up there makes a little less sense when the post it was responding to vanished...)

 

I don't understand. You understand. "Come let us reason together."

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Merged post follows:

Consecutive posts merged

Nicely done the tree.

 

Yeah!

[ajb]

(Apologies for the slight loss in continuity here when we removed some gibberish. ajb's post up there makes a little less sense when the post it was responding to vanished...)

 

Indeed.

 

Anyway, the tree gave us a nice way to attack this question. And then he showed that it does work.

[D H]

OK, so we're just showin' off?

 

The u-substitution [math]u = -\sqrt{x/4+1}[/math] works quite nicely here. With this substitution,

 

[math]\frac 1 {x\sqrt{x+4}}\,dx\,\rightarrow\,\frac{du}{1-u^2}[/math]

 

This integrates to tanh^-1u for u<1, coth^-1u for u>1, or more compactly,

 

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

= \frac 1 2 \ln \left|\frac{u+1}{u-1}\right|[/math]

 

Undoing the u-substitution,

 

[math]\int\frac 1 {x\sqrt{x+4}}\,dx

= \frac 1 2 \ln \frac {|\sqrt{x/4+1}-1|}{\sqrt{x/4+1}+1}

= \frac 1 2 \ln \frac {|\sqrt{x+4}-2|}{\sqrt{x+4}+2}[/math]

[dttom]

Thanks a lot, I got the idea now.