$\int \frac{dx}{x^{n+1} - x}\,dx$

Note: The $\pi$ in the original question is a typo — the integral is $\int \dfrac{dx}{x^{n+1} - x}$.

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Factor $x$ out of the denominator:

$x^{n+1} - x = x(x^n - 1)$

So the integral becomes:

$\int \dfrac{dx}{x(x^n - 1)}$

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Divide both the numerator and denominator by $x^{n+1}$ (i.e., multiply by $\dfrac{x^{-n-1}}{x^{-n-1}}$):

Denominator: $x(x^n - 1) \cdot x^{-n-1} = x^{-n}(x^n - 1) = 1 - x^{-n}$

Numerator: $1 \cdot x^{-n-1} = x^{-n-1}$

So we get:

$\int \dfrac{x^{-n-1}}{1 - x^{-n}}\,dx$

We do this because the derivative of the denominator will match the numerator — setting up a clean substitution.

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Let $u = 1 - x^{-n}$

$du = n\,x^{-n-1}\,dx$

$x^{-n-1}\,dx = \dfrac{du}{n}$

Substituting into the integral:

$\int \dfrac{1}{u} \cdot \dfrac{du}{n}$

$= \dfrac{1}{n} \int \dfrac{du}{u}$

$= \dfrac{1}{n} \ln|u| + C$

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Replace $u$ with $1 - x^{-n}$:

$= \dfrac{1}{n} \ln|1 - x^{-n}| + C$

Since $1 - x^{-n} = 1 - \dfrac{1}{x^n} = \dfrac{x^n - 1}{x^n}$, we can write:

$\int \dfrac{dx}{x^{n+1} - x} = \dfrac{1}{n} \ln\left|\dfrac{x^n - 1}{x^n}\right| + C$