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 :|
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 :|
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.
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.
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 $|x|<1$.