Evaluating $\int{ \frac{\arctan\sqrt{x^{2}-1}}{\sqrt{x^{2}+x}}} \,dx$ How to integrate?
$$\int{ \frac{\arctan\sqrt{x^{2}-1}}{\sqrt{x^{2}+x}}}\, dx$$
I have no idea how to do it.
Tried to get some information from wiki, but its too hard :|
 A: Putting $n=\sec\theta$ gives $$I=\int \frac{\theta \sec\theta \tan\theta}{\sec\theta\sqrt{\cos \theta+1}}\ d\theta=\int\frac{\theta\sin\frac{\theta}{2}}{\sqrt{\cos^2\frac{\theta}{2}+\frac{1}{2}}}\ d\theta$$ Substituting $\cos\theta/2$ by $x$ gives $$I=-2\int\frac{\cos^{-1}x}{\sqrt{x^2+a^2}}dx$$ where $a^2=1/2$. This does not seem to have any better form, at least as an indefinite integral.
A: Consider the integral 
$$\int\! \frac{\arccos x}{x\sqrt{x+1}} \, \mathrm{d}x,$$
which was derived by Mike. Using the property $\arccos x = \frac{\pi}{2}-\arcsin x$, we have
$$=\frac{\pi}{2}\int\! \frac{\mathrm{d}x}{x\sqrt{x+1}}-\int\! \frac{\arcsin x}{x\sqrt{x+1}} \, \mathrm{d}x.$$
The leftmost integral is evaluated trivially and so we consider the rightmost integral. We express $1/\sqrt{x+1}$ in terms of its MacLaurin series, valid for $|x|<1,$
$$\int \! \frac{\arcsin x}{x} \sum_{k=0}^{\infty} {-\frac{1}{2} \choose k} x^k \, \mathrm{d}x$$
$$=\sum_{k=0}^{\infty} {-\frac{1}{2} \choose k}\int\! x^{k-1}\arcsin x \, \mathrm{d}x.$$
Then, upon consideration of the integral
$$\int \! x^{k-1}\arcsin x\, \mathrm{d}x,$$
and an application of integration by parts, we find that
$$\int \! x^{k-1}\arcsin x\, \mathrm{d}x = \frac{x^k}{k}\arcsin x - \frac{1}{k}\int \! \frac{x^k \, \mathrm{d}x}{\sqrt{1-x^2}}. $$
According to Mathematica, the remaining integral may be expressed in terms of the hypergeometric function. This procedure may or may not provide an evaluation for $|x|<1$.
A: Mathematica returns a closed form solution for $\displaystyle\int\frac{\cos^{-1}x\,dx}{x\sqrt{x+1}}$, but it is several dozen lines long.  I am not sure how to clean it up yet.
