bincybn said:Hii Everyone,\(\displaystyle \int\frac{1}{x^{r}-1}dx
\) where r is a real no. greater than 1regards,
Bincy
chisigma said:If $|x|<1$ then is...
$\displaystyle \frac{1}{x^{r}-1} = - \sum_{n=0}^{\infty} x^{n\ r}$ (1)
... and You can integrate the expression (1) 'term by term'...
bincybn said:Can you please explain me the source of these formula?
Integrating $\frac{1}{x^r-1}$ with r>1 means finding the indefinite integral of the function $\frac{1}{x^r-1}$, where r is greater than 1. This involves finding a function whose derivative is equal to $\frac{1}{x^r-1}$.
The value of r determines the type of integral that must be used to integrate $\frac{1}{x^r-1}$. When r is greater than 1, the integral will be a logarithmic or inverse tangent integral.
No, $\frac{1}{x^r-1}$ cannot be integrated using basic integration techniques such as the power rule or substitution. It requires more advanced techniques such as partial fractions or trigonometric substitutions.
Integrating $\frac{1}{x^r-1}$ with r>1 has applications in various fields such as physics, engineering, and economics. It can be used to solve problems involving exponential growth and decay, as well as in the analysis of electrical circuits and financial markets.
Yes, there are limitations to integrating $\frac{1}{x^r-1}$ with r>1. In some cases, the integral may not have a closed form solution and must be evaluated numerically. Additionally, the integral may not exist for certain values of r, such as when r is equal to 1 or a negative integer.