Put ex = t
\(\frac {1}{(e^x-1) }\)
Let ex = t ⇒ ex dx = dt
⇒ \(∫\)\(\frac {1}{(e^x-1) }\) = \(∫\)\(\frac {1}{t-1}.\frac {dt}{t}\)= \(∫\)\(\frac {1}{t(t-1)} dt\)
Let \(\frac {1}{t(t-1)}\) = \(\frac {A}{t}+\frac {B}{t-1}\)
\(1 = A(t-1)+Bt\) ...(1)
Substituting t = 1 and t = 0 in equation (1), we obtain
\(A = −1 \ and \ B = 1\)
∴ \(\frac {1}{t(t-1)}\) = \(\frac {-1}{t}+\frac {1}{t-1}\)
⇒ \(∫\)\(\frac {1}{t(t-1)} dt\) = \(log|\frac {t-1}{t}|+C\)
= \(log\ |\frac {e^x-1}{e^x}|+C\)
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,
