\(\frac {5x}{(x+1)(x^2-4)}\)= \(\frac {5x}{(x+1)(x+2)(x-2)}\)
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 \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
If vector \( \mathbf{a} = 3 \hat{i} + 2 \hat{j} - \hat{k} \) \text{ and } \( \mathbf{b} = \hat{i} - \hat{j} + \hat{k} \), then which of the following is correct?
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,
