Evaluate the integral 

Correct Answer:

$\frac{\pi }{n}{\log }_e\left|\frac{x^n-1}{x^n}\right|+C$

Step-by-Step Explanation:

Let the given integral be:

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

First, we factor out the constant $\pi$ from the integral and factor out $x$ from the denominator:

$I=\pi \int \frac{1}{x\left(x^n-1\right)}dx$

To simplify this integral, we multiply both the numerator and the denominator by $x^{n-1}$:

$I=\pi \int \frac{x^{n-1}}{x^n\left(x^n-1\right)}dx$

Now, we can use the method of substitution. Let us define a new variable:

$t=x^n$

Differentiating both sides with respect to $x$ gives:

$dt=n{x}^{n-1}dx\Rightarrow x^{n-1}dx=\frac{dt}{n}$

Substituting $t$ and $dt$ into our integral:

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

We decompose the integrand using partial fractions:

$\frac{1}{t\left(t-1\right)}=\frac{t-\left(t-1\right)}{t\left(t-1\right)}=\frac{1}{t-1}-\frac{1}{t}$

Now, we integrate the separated terms:

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

$I=\frac{\pi }{n}\left({\log }_e\left|t-1\right|-{\log }_e\left|t\right|\right)+C$

Applying the logarithmic division property (${\log }_{e}A-{\log }_{e}B={\log }_e\frac{A}{B}$):

$I=\frac{\pi }{n}{\log }_e\left|\frac{t-1}{t}\right|+C$

Finally, substituting $t=x^n$ back into the expression yields the final answer:

$I=\frac{\pi }{n}{\log }_e\left|\frac{x^n-1}{x^n}\right|+C$