\(Let\ I=∫xsin\ 3xdx\)
Taking x as first function and \(sin\ 3x\) as second function and integrating by parts, we obtain
\(I= x∫sin\ 3x dx-∫{(\frac {d}{dx}x)∫sin\ 3x dx}\)
\(I = x(\frac {-cos\ 3x}{3})-∫1.(\frac {-cos\ 3x}{3})dx\)
\(I = \frac {-xcos\ 3x}{3}+\frac 13∫cos\ 3x dx\)
\(I = \frac {-xcos \ 3x}{3}+\frac 19sin\ 3x+C\)
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,
