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Find this integral $$\int\dfrac{\sqrt{x-1}\arctan{(x\sqrt{x-1})}}{x}dx$$

My try: let $$\arctan{(x\sqrt{x-1})}=t$$ and that's very ugly,Thank you

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Mathematica can't find a primitive at all. – Lucian Dec 2 '13 at 10:09
I would really like to learn how to compute this antiderivative ! – Claude Leibovici Dec 2 '13 at 10:13

$\because$ according to, $\int\dfrac{\sqrt{x-1}}{x}dx=2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1}+C$




$\because$ according to, $\dfrac{d}{dx}(\tan^{-1}(x\sqrt{x-1}))=\dfrac{3x-2}{2\sqrt{x-1}(x^3-x^2+1)}$




$\int\dfrac{3x-2}{x^3-x^2+1}dx$ is just an integral of rational functions and should have close form.

For $\int\dfrac{(3x-2)\tan^{-1}\sqrt{x-1}}{\sqrt{x-1}(x^3-x^2+1)}dx$ ,

Let $u=\sqrt{x-1}$ ,

Then $x=u^2+1$




Then you can separate it to the terms of $\int\dfrac{\tan^{-1}u}{au+b}du$ or $\int\dfrac{\tan^{-1}u}{pu^2+qu+r}du$ by partial fraction. According to, they relate to the polylogarithm function.

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this is unreadable – Norbert Dec 5 '13 at 17:06
Even that integral of rational function is not elementary, so (-1) – Norbert Dec 5 '13 at 17:15
@Norbert: I don't agree with you. For partial fraction using in integration, it is not a must to factorize in $\mathbb{Q}$ , factorize in $\mathbb{R}$ or even in $\mathbb{C}$ is also allowed. The downvotes should be unreasonable. – Harry Peter Dec 9 '13 at 14:15
Then please provide us with explicit formula for the antiderivative – Norbert Dec 9 '13 at 14:41

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