We are asked to evaluate: \[ \int \frac{1}{x(x^4 + 1)} \, dx \] We will use partial fraction decomposition to simplify the integrand. The integrand can be expressed as: \[ \frac{1}{x(x^4 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^4 + 1} \] After performing partial fraction decomposition (details omitted for brevity), we integrate each term. The result of this integral is: \[ \frac{1}{2} \log |x^4 + 1| + C \]
Thus, the correct answer is \( \frac{1}{2} \log |x^4 + 1| \).
Let $ f(x) + 2f\left( \frac{1}{x} \right) = x^2 + 5 $ and $ 2g(x) - 3g\left( \frac{1}{2} \right) = x, \, x>0. \, \text{If} \, \alpha = \int_{1}^{2} f(x) \, dx, \, \beta = \int_{1}^{2} g(x) \, dx, \text{ then the value of } 9\alpha + \beta \text{ is:}$