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 \[ f(t)=\int \left(\frac{1-\sin(\log_e t)}{1-\cos(\log_e t)}\right)dt,\; t>1. \] If $f(e^{\pi/2})=-e^{\pi/2}$ and $f(e^{\pi/4})=\alpha e^{\pi/4}$, then $\alpha$ equals