rational integral

**waytogo** · June 10th 2010, 12:06 AM

Hi, can somebody give any hint how to get this integral $\int\frac{x^3}{x^4+x^3+x^2+x+1}dx$ ?

**chisigma** · June 10th 2010, 02:30 AM

Is...

$1 + x + x^{2} + x^{3} + x^{4} = \prod_{k=1}^{4} (x-x_{k})$ , $x_{k} = e^{\frac{2 \pi k i}{5}}$ (1)

... so that...

$\frac{x^{3}}{1 + x + x^{2} + x^{3} + x^{4}}= \sum_{k=1}^{4} \frac{r_{k}}{(x-x_{k})}$ (2)

... where...

$r_{k} = \lim_{x \rightarrow x_{k}} \frac {x^{3}\cdot (x-x_{k})}{1 + x + x^{2} + x^{3} + x^{4}} = \frac{x_{k}^{3}}{1 + 2 x_{k} + 3 x_{k}^{2} + 4 x_{k}^{3}}$ (3)

... so that is...

$\int \frac{x^{3}}{1 + x + x^{2} + x^{3} + x^{4}} dx = \sum_{k=1}^{4} r_{k} \cdot \ln (x-x_{k}) + c$ (4)

Kind regards

$\chi$ $\sigma$

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