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## Rational Integral

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### Rational Integral

Fri Dec 30, 2016 7:32 am

Posts: 82
If $\displaystyle \displaystyle g(x) = \int \frac{x^2-12}{(x^2-6x+k)^2}dx ,(k\in \mathbb{N})$ is a rational function. Then find possible values of $\displaystyle k$

### Re: Rational Integral

Fri Dec 30, 2016 9:16 am

Posts: 22
If $k<9$,

$g(x)=\frac{\left( k-12\right) \cdot \mathrm{log}\left( \frac{-6-2\cdot \sqrt{9-k}+2\cdot x}{2\cdot x+2\cdot \sqrt{9-k}-6}\right) }{2\cdot \sqrt{9-k}\cdot \left( 2\cdot k-18\right) }-\frac{-36+3\cdot k+\left( k-6\right) \cdot x}{\left( 2\cdot k-18\right) \cdot {{x}^{2}}+\left( 108-12\cdot k\right) \cdot x+2\cdot {{k}^{2}}-18\cdot k}$

If $k>9$,

$g(x)=\frac{\left( k-12\right) \cdot \mathrm{atan}\left( \frac{2\cdot x-6}{2\cdot \sqrt{k-9}}\right) }{2\cdot {{\left( k-9\right) }^{\frac{3}{2}}}}-\frac{-36+3\cdot k+\left( k-6\right) \cdot x}{\left( 2\cdot k-18\right) \cdot {{x}^{2}}+\left( 108-12\cdot k\right) \cdot x+2\cdot {{k}^{2}}-18\cdot k}$

If $k$ isn't $9$, $g(x)$ isn't a rational function. (not really a proof i know).

If $k=9$,

$g(x)=-\frac{-1-3\cdot x+{{x}^{2}}}{{{x}^{3}}-9\cdot {{x}^{2}}+27\cdot x-27}$