Nightmare integral

Hi all,
I need ur help with this integral
[tex]\int\frac{x*ln(x+sqrt{1+x^2}*dx} {sqrt{1+x^2} [/tex]
maybe its about substitution but how?
 
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arildno

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Vereinsamt said:
Hi all,
I need ur help with this integral
[tex]\int\frac{xln(x+\sqrt{1+x^2})} {\sqrt{1+x^2}}dx[/tex]
maybe its about substitution but how?
Did you mean that?
 
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exactly :)
 

arildno

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At first glance, integration by parts seems best, with [tex]\frac{dv}{dx}=\frac{x}{\sqrt{1+x^{2}}}[/tex]
 

Hurkyl

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It seems to me that there are two obvious substitutions to try. Have you tried anything yet?

Not to mention IBP -- you've surely seen how to do the integrals of x ln x and ln x by now, and seen the theme there?
 
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arildno said:
At first glance, integration by parts seems best, with [tex]\frac{dv}{dx}=\frac{x}{\sqrt{1+x^{2}}}[/tex]
by doing this I got

[tex]\int{udv}=ln(x+t)t-\int\frac{(2t+1)xdx} {x+t}[/tex]
where [tex]t=\sqrt{1+x^2}[/tex]

is it right? and how then?
 
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arildno

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Incorrect!
We have:
[tex]\frac{dt}{dx}=\frac{x}{t}[/tex]
Thus, your last integral is:
[tex]\int\frac{t(1+\frac{x}{t})}{x+t}dx=\int{dx}[/tex]
 
Hurkyl said:
It seems to me that there are two obvious substitutions to try. Have you tried anything yet?

Not to mention IBP -- you've surely seen how to do the integrals of x ln x and ln x by now, and seen the theme there?
I am a self-study and new to calculus and have no experience, so I hope u explain more about ur point
 

Hurkyl

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I am a self-study and new to calculus and have no experience, so I hope u explain more about ur point
One of them relies on trigonometric substitution -- I suppose if you haven't gotten that far, then the relevant substitution would not be obvious. :smile:


The other one stems from the fact you (presumably) know how to integrate things that look like "ln x", but not "ln [something other than a plain ordinary x]". So, one obvious thing to try is to do something that will turn "ln [something other than a plain ordinary x]" into something akin to "ln x".
 
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arildno said:
Incorrect!
We have:
[tex]\frac{dt}{dx}=\frac{x}{t}[/tex]
Thus, your last integral is:
[tex]\int\frac{t(1+\frac{x}{t})}{x+t}dx=\int{dx}[/tex]
yes. there was an error in deffirentiating ln.
I appiciate ur help arildno :)
anyway what about other interesting methods?

thanx alot
 

arildno

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Well, you might try the substitution x=Sinh(u), where Sinh() is the hyperbolic sine function.
 

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