# Thread: How to integrate sqrt(arctanx)?

1. ## How to integrate sqrt(arctanx)?

This question is posed here: How to integrate sqrt(arctanx)? - Yahoo! Answers

I think the presumption of the only answer by Robert V is correct and that the OP got the order of operation mixed up.

However, maybe someone here can provide a closed form answer to the original question?

(If not, then thanks for your efforts.)

2. ## Re: How to integrate sqrt(arctanx)?

Originally Posted by NowIsForever
This question is posed here: How to integrate sqrt(arctanx)? - Yahoo! Answers

I think the presumption of the only answer by Robert V is correct and that the OP got the order of operation mixed up.

However, maybe someone here can provide a closed form answer to the original question?

(If not, then thanks for your efforts.)
For:
$\int \tan^{-1}(\sqrt{x})\,\,dx$

Let $s = \sqrt{x} \therefore ds = \frac{1}{2\sqrt{x}}\,\,dx = \frac{1}{2s}\,\,dx$

So $dx = 2s\,\,ds$

\begin{align*}\int \tan^{-1}{(\sqrt{x})}\,\,dx = \int 2s\tan^{-1}{(s)}\,\,ds....\text{[from above]}\end{align*}

$\text{Let } u = \tan^{-1}{(s)} \therefore du = \frac{1}{1+s^2}\,\, ds$

$\text{And } dv = 2s\,\, ds \therefore v = s^2$

Now using integration by parts:

\begin{align*}&\int 2s\tan^{-1}{(s)}\,\,ds \\=& s^{2}\tan^{-1}{(s)} - \int \frac{s^2}{1+s^2}\,\,ds\\=& s^{2}\tan^{-1}{(s)} - \int \left(1 - \frac{1}{1+s^2}\right)\,\, ds\\=& s^{2}\tan^{-1}{(s)} - \int \,\,ds + \int \frac{1}{1+s^2}\,\,ds\\=& s^{2}\tan^{-1}{(s)} - s + \tan^{-1}(s) +C \\ =& x\tan^{-1}{(\sqrt{x})} - \sqrt{x} + \tan^{-1}(\sqrt{x}) + C.....\text{[because }s = \sqrt{x}\text{]}\end{align*}

You can check if my answer is correct here:
int arctan((x)^(1/2)) - Wolfram|Alpha

Hope this helps.

3. ## Re: How to integrate sqrt(arctanx)?

[QUOTE

Hope this helps.[/QUOTE]

4. ## Re: How to integrate sqrt(arctanx)?

Thanks, x3bnm, but you have answered the reverse question just as Robert V did in Yahoo! Answers. The original (albeit, possibly mistaken) question was int sqrt(arctan(x))?

wolframalpha.com couldn't do it either, though perhaps the Pro version could.

I suspect there is no closed form solution, although proving that could be incredibly difficult.

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