A question on integration I want to compute the following integral:
$$\raise 1ex{\Large\int} \frac{\sqrt{\ln(x+\sqrt{1+x^2}})}{1+x^2}\,dx$$
 A: Why not make the substitution $x=\sinh u?$ Then, $d x = \cosh u d u,$ while $(x+\sqrt{1+x^2})= \exp(u),$ and $1+x^2 = \cosh^2 u,$ so we get our integral equal to 
$$\int \frac{\sqrt u}{\cosh u} du,$$ Now, letting $u = v^2,$ this is equal to
$$\frac12\int \operatorname{sech}(v^2) d v.$$ At which point I am flummoxed.
A: I'd add this as a comment but I don't have enough reputation to do so.  Building off Igor's work, I wondered whether $\frac{2v^2}{\cosh(v^2)}$ could be done by parts.  If we let $f=v$ and $g'=2v\text{sech}(v^2)$, then $fg-\int{f'g}$ works out to $$2v\arctan(e^{v^2})-2\int{\arctan(e^{v^2})dv}$$
Of course, I had to go to Wolfram for the second half:
$$ 2 \int{\arctan(e^{v^2}) dv} = \sqrt{\frac{\pi}{\ln(\arctan(e))}}\text{erfi}\left(v\sqrt{\ln(\arctan(e))}\right)$$
where $\text{erfi}$ is apparently the imaginary error function.  (I know a little about the plain ol' error function, but...)  Unless there's something weird with that error function thing, I'd assume we can just undo the substitution, with $v=\sqrt{u}=\sqrt{\text{arcsinh}(x)}$:
$$2\sqrt{\text{arcsinh}(x)}\arctan(e^{\text{arcsinh}(x)})-\sqrt{\frac{\pi}{\ln(\arctan(e))}}\text{erfi}\sqrt{\text{arcsinh}(x)\ln(\arctan(e))}$$
I'm guessing, though, that most of the people looking at this know what they're doing more than I do, so I wouldn't be surprised if there's some reason we can't do the above.
I'm trying to check this on Wolfram, just by subtracting the original integrand from the derivative of the above, but it's not interpreting the parentheses correctly.  This is what I'm entering:
d/dx{
2sqrt{arcsinh(x)}arctan(e^{arcsinh(x)})-sqrt{pi/ln(arctan(e))}*erfi(sqrt{arcsinh(x)ln(arctan(e))})
}
-sqrt(ln(x+sqrt(x^2+1)))/(1+x^2)
It keeps taking the derivative of the whole thing.  I'm pretty sure I parsed that carefully, but I thought I'd include it in case anyone else wants to try.
