\(\frac {e^x}{(1+e^x)(2+e^x)}\)
\(Let\ e^x=t ⇒ e^x dx=dt\)
\(⇒\)∫\(\frac {e^x}{(1+e^x)(2+e^x)}dx\) = \(∫\frac {dt}{(t+1)(t+2)}\)
=\(∫[\frac {1}{(t+1)}-\frac {1}{(t+2)}]dt\)
= \(log\ |t+1|-log\ |t+2|+C\)
= \(log|\frac {t+1}{t+2}|+C\)
= \(log|\frac {1+e^x}{2+e^x}|+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
The number of formulas used to decompose the given improper rational functions is given below. By using the given expressions, we can quickly write the integrand as a sum of proper rational functions.

For examples,
