How to integrate $$\int_1^{\infty}\frac{2x^3-1}{x^6+2x^3+\sqrt3x^2+1}{\rm d}x$$ The bottom is not factorizable hence no partial fractions. There seems no other way.
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4A non-factorizable 6th-degree polynomial? I don't think so. – GEdgar Apr 13 '15 at 12:42
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1@GEdgar ಠ_ಠ ${}$ – RE60K Apr 13 '15 at 12:52
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$$\begin{align}\int_{1}^{\infty}\frac{2x^3-1}{x^6+2x^3+\sqrt 3x^2+1}dx&=\int_{1}^{\infty}\frac{2x^3-1}{(x^3+1)^2+\sqrt 3x^2}dx\\&=\int_{1}^{\infty}\frac{(2x^3-1)/(x^2)}{\left((x^3+1)^2+\sqrt 3x^2\right)/(x^2)}dx\\&=\int_{1}^{\infty}\frac{2x-\frac{1}{x^2}}{\left(x^2+\frac 1x\right)^2+\sqrt 3}dx\\&=\int_{2}^{\infty}\frac{du}{u^2+\sqrt 3}\ \ \ \ (\text{set $x^2+\frac 1x=u$})\\&=\int_{\arctan\frac{2}{\sqrt[4]{3}}}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt[4]{3}}\ \ \ \ (\text{set $u=\sqrt[4]{3}\tan\theta$})\\&=\frac{1}{2\sqrt[4]{3}}\left(\pi-2\arctan\frac{2}{\sqrt[4]{3}}\right).\end{align}$$
mathlove
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@ADG: All I noticed was $x^6+2x^3+1=(x^3+1)^2$. After that, only one way, isn't it? – mathlove Apr 13 '15 at 12:16
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@ADG: You'll get $\int_{2}^{\infty}\frac{du}{u^2+\sqrt 3}$. I'm sure you can continue. – mathlove Apr 13 '15 at 12:32
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yes I can but i just asked if you would comlete here for the sake of completenes, anyways you say, arctan u/sqrtsqrt3. – RE60K Apr 13 '15 at 12:36
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