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?

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

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.

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, 201015 yr by the tree

Nicely done the tree.

(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...)

(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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Nicely done the tree.

 

Yeah!

(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.

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]

Thanks a lot, I got the idea now.