I misread an integral on my school maths book coming up with this indefinite integral $$\int x \sqrt{x-1}\arctan x\, \text{d}x$$ that revealed to be quite nasty. The first thing I thought has been to substitute $x-1=t^2$ and go by parts deriving $\arctan x$ obtaining $$\frac2{15}\left((3t^5+5t^3)\arctan (t^2+1)-2\int\frac{3t^6+5t^4}{t^4+2t^2+2}\, \text{d}t \right)$$ The problem now is that rational integral that looks even worse than the original one. In fact, throught sinthtic division the integrand becomes $3t^2-1-2\frac{2t^2-1}{t^4+2t^2+2}$ that still keeps the quartic denominator with no chance to express the numerator as the derivative or to apply add-and-subtract something… Then, I think that my initial substitution isn't the best but also trying with $x=\tan u$ I have no great results as well as with something more. Any idea?
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Just out of curiosity, what was the actual integral on your school book? – Toby Mak Apr 09 '20 at 10:48
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Way too simpler ahah! However, that is $\int x\arctan\sqrt{x-1},\text{d}x$ – bianco Apr 09 '20 at 11:01
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What do you mean ? Is, as posted, the integrand wrong ? – Claude Leibovici Apr 09 '20 at 11:03
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1I was doing integral by parts exercises on my school book and instead of reading $\int x\arctan\sqrt{x-1},\text{d}x$ I read the one I posted about – bianco Apr 09 '20 at 11:10
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Hint. In the last rational function, the denominator is quadratic in $t^2.$ So you can factor into two quadratics and express the fraction in parts. This would require manipulating complex entities, but the procedures are the same.
Another way may be to try a substitution of the form $$x=at+\frac bt,$$ where $a$ and $b$ are suitable constants.
Allawonder
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@MaximilianJanisch That's clearly the case, but by complex I wasn't referring to that. I meant it would involve dealing with complex numbers, which OP might not be cool with. Hence my suggestion of the substitution. – Allawonder Apr 09 '20 at 11:36
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Before posting I tried to go with complex numbers but I got many mistakes in the system for decomposing the denominator. Maybe now I have a solution but it's absurd! That is $\frac2{15}\left((3x-2)\sqrt{x-1}^3\arctan x-2\sqrt{x-1}(x-2)+2\frac{3i+2}{z}\ln\left|\frac{\sqrt{x-1}-z}{\sqrt{x-1}+z}\right|-2\frac{3i-2}{\bar{z}}\ln\left|\frac{\sqrt{x-1}-\bar{z}}{\sqrt{x-1}+\bar{z}}\right|\right)+C$ where $z=\sqrt[4]{2}e^{i\frac{3\pi}8}$. Could you explain how to do with that substitution? – bianco Apr 09 '20 at 16:40
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@bianco Make the substitution, then after representing the integral in terms of $a,b$ determine the values of $a,b$ so that the integral is in as simple a form as you wish. – Allawonder Apr 09 '20 at 18:43