Solve Integral: Nightmare | x*ln(x+sqrt{1+x^2})

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Homework Help Overview

The discussion revolves around the integral \(\int\frac{x \ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx\), with participants exploring various methods of integration, including substitution and integration by parts.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants suggest integration by parts and discuss potential substitutions, including trigonometric and hyperbolic functions. There is a focus on how to manipulate the logarithmic expression within the integral.

Discussion Status

The conversation is ongoing, with participants sharing different approaches and questioning the validity of each other's methods. Some guidance has been offered regarding integration techniques, but no consensus has been reached on a specific solution.

Contextual Notes

Some participants express their lack of experience in calculus, which may influence their understanding of the suggested methods and substitutions.

Vereinsamt
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Hi all,
I need ur help with this integral
\int\frac{x*ln(x+sqrt{1+x^2}*dx} {sqrt{1+x^2}
maybe its about substitution but how?
 
Last edited:
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Vereinsamt said:
Hi all,
I need ur help with this integral
\int\frac{xln(x+\sqrt{1+x^2})} {\sqrt{1+x^2}}dx
maybe its about substitution but how?
Did you mean that?
 
Last edited:
exactly :)
 
At first glance, integration by parts seems best, with \frac{dv}{dx}=\frac{x}{\sqrt{1+x^{2}}}
 
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 \frac{dv}{dx}=\frac{x}{\sqrt{1+x^{2}}}

by doing this I got

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

is it right? and how then?
 
Last edited:
Incorrect!
We have:
\frac{dt}{dx}=\frac{x}{t}
Thus, your last integral is:
\int\frac{t(1+\frac{x}{t})}{x+t}dx=\int{dx}
 
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
 
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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  • #10
arildno said:
Incorrect!
We have:
\frac{dt}{dx}=\frac{x}{t}
Thus, your last integral is:
\int\frac{t(1+\frac{x}{t})}{x+t}dx=\int{dx}

yes. there was an error in deffirentiating ln.
I appiciate ur help arildno :)
anyway what about other interesting methods?

thanx a lot
 
  • #11
Well, you might try the substitution x=Sinh(u), where Sinh() is the hyperbolic sine function.
 

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