$\int \dfrac{\pi}{x^{n+1}-x} d x=$

We need to evaluate $\int \dfrac{\pi}{x^{n+1}-x} dx$

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First, let's rewrite the denominator:

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

So our integral becomes:

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

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Using partial fractions:

$\dfrac{\pi}{x(x^n-1)} = \dfrac{A}{x} + \dfrac{B}{x^n-1}$

Multiplying all terms by $x(x^n-1)$:

$\pi = A(x^n-1) + Bx$

When $x = 0$: $\pi = A(-1)$, so $A = -\pi$

When $x = 1$: $\pi = B$, so $B = \pi$

Therefore:

$\dfrac{\pi}{x^{n+1}-x} = \dfrac{-\pi}{x} + \dfrac{\pi}{x^n-1}$

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Integrating each term:

$\int \dfrac{\pi}{x^{n+1}-x}dx = \int \dfrac{-\pi}{x}dx + \int \dfrac{\pi}{x^n-1}dx$

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

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For the second integral, use substitution $t = x^n$:

$dt = nx^{n-1}dx$ → $dx = \dfrac{dt}{nx^{n-1}}$

$\int \dfrac{1}{x^n-1}dx = \int \dfrac{1}{t-1} \cdot \dfrac{dt}{nx^{n-1}}$

Since $x^{n-1} = \dfrac{t}{x}$, we have:

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

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Let's try direct substitution with $t = x^n$:

$\int \dfrac{\pi}{x(x^n-1)}dx = \dfrac{\pi}{n} \int \dfrac{dt}{t(t-1)}$

$= \dfrac{\pi}{n} \int \left(\dfrac{1}{t-1} - \dfrac{1}{t}\right)dt$

$= \dfrac{\pi}{n}[\ln|t-1| - \ln|t|] + C$

$= \dfrac{\pi}{n}\ln\left|\dfrac{t-1}{t}\right| + C$

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Substituting back $t = x^n$:

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