# Nightmare integral

#### 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?

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#### arildno

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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?

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exactly :)

#### arildno

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

#### 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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#### Vereinsamt

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?

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#### arildno

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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

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. 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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#### Vereinsamt

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 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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