Originally Posted by

**Revy** I'm trying to integrate $\displaystyle \int\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx$

I start with

$\displaystyle u = \ln(x+\sqrt{1+x^2})$

$\displaystyle \frac{du}{dx}=\frac{1}{x}(1+\frac{x}{\sqrt{1+x^2}} )=\frac{1}{x}+\frac{1}{\sqrt{1+x^2}}$

Can I do like this? $\displaystyle \ln{x} + \ln{\sqrt{1+x^2}}=\ln(x+\sqrt{1+x^2})$

if so, then do I continue like this? $\displaystyle \int{2u} du= u^2$

Hm... and I noticed that I rewrited $\displaystyle \ln$ backward xD omg, can anyone help me out here?