Evaluate the integral using the substitution \( u=x^{5}+5 x \) or otherwise \( \int_{2}^{3} \frac{x^{4}+1}{x^{5}+5 x} d x \)

To evaluate the integral \( I = \int_{2}^{3} \frac{x^{4}+1}{x^{5}+5 x} \, dx \) using the substitution \( u = x^{5} + 5x \), we first find the derivative of \( u \): \[ \frac{du}{dx} = 5x^4 + 5. \] This simplifies to: \[ du = (5x^4 + 5) \, dx = 5(x^4 + 1) \, dx. \] From this, we find: \[ dx = \frac{du}{5(x^4 + 1)}. \] Now, we also need to change the limits of integration. When \( x = 2 \): \[ u(2) = 2^5 + 5 \cdot 2 = 32 + 10 = 42, \] and when \( x = 3 \): \[ u(3) = 3^5 + 5 \cdot 3 = 243 + 15 = 258. \] Substituting these into the integral gives: \[ I = \int_{42}^{258} \frac{x^{4}+1}{u} \cdot \frac{du}{5(x^{4} + 1)}. \] The \( x^{4} + 1 \) in the numerator and denominator cancels out, leading to: \[ I = \frac{1}{5} \int_{42}^{258} \frac{1}{u} \, du. \] Now, we can evaluate the integral: \[ \int \frac{1}{u} \, du = \ln |u| + C. \] So, we have: \[ I = \frac{1}{5} \left[ \ln |u| \right]_{42}^{258} = \frac{1}{5} \left( \ln(258) - \ln(42) \right) = \frac{1}{5} \ln\left(\frac{258}{42}\right). \] Thus, the final result of the integral is: \[ \boxed{ \frac{1}{5} \ln\left(\frac{258}{42}\right) }. \]