Let \(\frac{x}{(x+1)(x-+2)} = \frac{A}{(x+1)}+\frac{B}{(x+2)}\)
\(\Rightarrow x = A(x+2)+B(x+1)\)
Equating the coefficients of x and constant term, we obtain
A + B = 1
2A + B = 0
On solving, we obtain
A = −1 and B = 2
∴ \(\frac{x}{(x+1)(x+2)}=\frac{-1}{(x+1)}+\frac{2}{(x+2)}\)
\(\Rightarrow \int \frac{x}{(x+1)(x+2)}dx = \int \frac{-1}{(x+1)}+\frac{2}{(x+2)}dx\)
= \(- \log \mid x+1 \mid +2\log \mid x+2 \mid+C\)
= \(\log(x+2)^2 -\log \mid x+1 \mid+C\)
= \(\log(\frac{x+2)^2}{ (x+1)}+C\)
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,