juantheron
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$(1)\displaystyle \;\; \int\frac{1}{1+x^6}dx$
$(2)\displaystyle \;\; \int\frac{1}{1+x^8}dx$
$(2)\displaystyle \;\; \int\frac{1}{1+x^8}dx$
The integration of the function \( \frac{1}{1+x^n} \) can be effectively approached using the roots of the polynomial \( 1+x^n \). The roots are given by \( x_k = e^{i \frac{2k+1}{n} \pi} \) for \( k = 0, 1, \ldots, n-1 \). By expressing \( \frac{1}{1+x^n} \) as a sum of partial fractions, \( \frac{1}{1+x^n} = \sum_{k=0}^{n-1} \frac{r_k}{x-x_k} \), and integrating term by term, the result is \( \int \frac{dx}{1+x^n} = \sum_{k=0}^{n-1} r_k \ln(x-x_k) + c \).
PREREQUISITESMathematicians, calculus students, and anyone interested in advanced integration techniques and the application of polynomial roots in mathematical analysis.
jacks said:$(1)\displaystyle \;\; \int\frac{1}{1+x^6}dx$
$(2)\displaystyle \;\; \int\frac{1}{1+x^8}dx$