# Nightmare integral

1. Feb 16, 2006

### 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?

Last edited: Feb 16, 2006
2. Feb 16, 2006

### arildno

Did you mean that?

Last edited: Feb 16, 2006
3. Feb 16, 2006

### Vereinsamt

exactly :)

4. Feb 16, 2006

### arildno

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

5. Feb 16, 2006

### Hurkyl

Staff Emeritus
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?

Last edited: Feb 16, 2006
6. Feb 16, 2006

### Vereinsamt

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: Feb 16, 2006
7. Feb 16, 2006

### arildno

Incorrect!
We have:
$$\frac{dt}{dx}=\frac{x}{t}$$
$$\int\frac{t(1+\frac{x}{t})}{x+t}dx=\int{dx}$$

8. Feb 16, 2006

### Vereinsamt

I am a self-study and new to calculus and have no experience, so I hope u explain more about ur point

9. Feb 16, 2006

### Hurkyl

Staff Emeritus
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".

Last edited: Feb 16, 2006
10. Feb 16, 2006

### Vereinsamt

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

thanx alot

11. Feb 16, 2006

### arildno

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