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2 - Indefinite Integrals

jacks

Well-known member
Apr 5, 2012
226
$(1)\displaystyle \;\; \int\frac{1}{1+x^6}dx$

$(2)\displaystyle \;\; \int\frac{1}{1+x^8}dx$
 

chisigma

Well-known member
Feb 13, 2012
1,704
$(1)\displaystyle \;\; \int\frac{1}{1+x^6}dx$

$(2)\displaystyle \;\; \int\frac{1}{1+x^8}dx$
May be is useful to consider the general case...

$\displaystyle \int \frac{dx}{1+x^{n}}$ (1)

The root of the polinomial $\displaystyle 1+x^{n}$ are $\displaystyle x_{k}= e^{i\ \frac{2k+1}{n}\ \pi}\ ,\ k= 0.1,...,n-1$ so that is...

$\displaystyle \frac{1}{1+x^{n}} = \sum_{k=0}^{n-1} \frac{r_{k}}{x-x_{k}}$ (2)

... where...

$\displaystyle r_{k}= \lim_{x \rightarrow x_{k}} \frac{x-x_{k}}{1+x^{n}}$ (3)

Now You can integrate (2) 'term by term' obtaining...


$\displaystyle \int \frac{dx}{1+x^{n}}= \sum_{k=0}^{n-1} r_{k}\ \ln (x-x_{k}) + c $ (4)

Kind regards

$\chi$ $\sigma$
 
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