The given integral is: \[ \int e^{-x} \cdot e^{3x} \, dx \] Using the property of exponents that \( e^a \cdot e^b = e^{a + b} \), we can combine the exponents: \[ \int e^{-x + 3x} \, dx = \int e^{2x} \, dx \] Now, integrate \( e^{2x} \) with respect to \( x \): \[ \int e^{2x} \, dx = \frac{e^{2x}}{2} + C \]
Thus, the correct answer is \( \frac{e^{2x}}{2} + C \).
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