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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 :|

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Please try to use a more specific title, this one is very generic and will not assist users searching for similar queries in future. –  Simon Hayward Dec 1 '12 at 21:10
It's easier for people if you use conventional letters for variables. Since only one variable appears in your question, it would be more friendly to call it $x$. The letter $n$ normally denotes an integer. –  John Bentin Dec 1 '12 at 21:24
The substitution $n=\sec\theta$ looks messy. I've also gotten it into the form $\int\frac{\cos^{-1}xdx}{x\sqrt{x+1}}$ and tried integration by parts, but that doesn't seem to lead anywhere as eliminating the inverse trig function still leaves natural log terms... –  Mike Dec 2 '12 at 18:16
@user51402 looks like Wolfram Mathematica refuses symbolic integration. –  Caran-d'Ache May 18 '13 at 8:19
Hope you are not randomly writing integrands and asking us. And now I realize I just brought up an old question... –  evil999man Mar 26 '14 at 14:20

3 Answers 3

up vote 1 down vote accepted

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 $|x|<1$.

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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.

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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.

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The solution you speak of can be reduced to the size of a single screen (six or seven lines) if FullSimplify is employed. –  Lucian Oct 31 '13 at 2:07
Please show the amazing result. –  doraemonpaul Feb 9 '14 at 0:02

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