dttom Posted April 30, 2010 Share Posted April 30, 2010 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? Link to comment Share on other sites More sharing options...
the tree Posted April 30, 2010 Share Posted April 30, 2010 Maybe you could start by working out [imath]\frac{\mbox{d}}{\mbox{d}x} \ln(\sqrt{x+a}+b)[/imath]. Link to comment Share on other sites More sharing options...
ajb Posted May 4, 2010 Share Posted May 4, 2010 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. Link to comment Share on other sites More sharing options...
the tree Posted May 4, 2010 Share Posted May 4, 2010 (edited) 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 Link to comment Share on other sites More sharing options...
ajb Posted May 4, 2010 Share Posted May 4, 2010 Nicely done the tree. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted May 4, 2010 Share Posted May 4, 2010 (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...) Link to comment Share on other sites More sharing options...
Charlatan Posted May 4, 2010 Share Posted May 4, 2010 (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." Merged post follows: Consecutive posts mergedNicely done the tree. Yeah! Link to comment Share on other sites More sharing options...
ajb Posted May 4, 2010 Share Posted May 4, 2010 (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. Link to comment Share on other sites More sharing options...
D H Posted May 4, 2010 Share Posted May 4, 2010 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] Link to comment Share on other sites More sharing options...
dttom Posted May 5, 2010 Author Share Posted May 5, 2010 Thanks a lot, I got the idea now. Link to comment Share on other sites More sharing options...
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