\(\frac {5x}{(x+1)(x^2-4)}\)= \(\frac {5x}{(x+1)(x+2)(x-2)}\)
Let \(\frac {5x}{(x+1)(x+2)(x-2)}\) = \(\frac {A}{(x+1)} +\frac {B}{(x+2)} + \frac {C}{(x-2)}\)
\(5x = A(x+2)(x-2) + B(x+1)(x-2) + C(x+1)(x+2) ….....(1)\)
Substituting x = −1, −2, and 2 respectively in equation (1), we obtain
A = \(\frac 53\), B= \(-\frac {5}{2}\), and C = \(\frac 56\)
∴ \(\frac {5x}{(x+1)(x+2)(x-2)}\) = \(\frac {5}{3(x+1)} -\frac {5}{2(x+2)} + \frac {5}{6(x-2)}\)
⇒ \(∫\)\(\frac {5x}{(x+1)(x+2)(x-2)} \ dx\) = \(\frac 53 ∫\frac {1}{(x+1)}dx-\frac 52 ∫\frac {1}{(x+2)}dx+\frac 56 ∫\frac {1}{(x-2)}dx\)
= \(\frac 53\ log|x+1|-\frac 52\ log|x+2|+\frac 56\ log|x-2|+C\)
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,
