# Integration

#### bincybn

##### Member
Hii Everyone,

$$\displaystyle \int\frac{1}{x^{r}-1}dx$$ where r is a real no. greater than 1

regards,
Bincy

#### Sudharaka

##### Well-known member
MHB Math Helper
Hii Everyone,

$$\displaystyle \int\frac{1}{x^{r}-1}dx$$ where r is a real no. greater than 1

regards,
Bincy
Hi Bincy,

This integral may not be expressible in terms of elementary functions. See this.

Kind Regards,
Sudharaka.

#### chisigma

##### Well-known member
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'...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
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'...
... and if $|x|>1$ then You can set $\displaystyle t=\frac{1}{x}$ and the function to be integrated becomes...

$\displaystyle \frac{1}{t^{2}\ \{1-(\frac{1}{t})^{r}\}}\ = - \sum_{n=0}^{\infty} t^{(n+1)\ r-2}$ (1)

... and also in this case You can integrate 'term by term'...

Kind regards

$\chi$ $\sigma$

#### bincybn

##### Member
Can you plz explain me the source of these formula?

#### CaptainBlack

##### Well-known member
Can you plz explain me the source of these formula?
chisigma is just using the sum of an infinite geometric series:

$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}, \ \ \text{for } |x|<1$

which should be common knowledge.

CB