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

[Vereinsamt]

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?

 

[arildno]

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?

 

[Vereinsamt]

exactly :)

 

[arildno]

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

 

[Hurkyl]

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?

 

[Vereinsamt]

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?

 

[arildno]

Incorrect!
We have:
$$\frac{dt}{dx}=\frac{x}{t}$$
Thus, your last integral is:
$$\int\frac{t(1+\frac{x}{t})}{x+t}dx=\int{dx}$$

 

[Vereinsamt]

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]

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.


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".

 

[Vereinsamt]

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

 

[arildno]

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