Let I = ∫x2ex dx
Taking \(x^2\) as first function and ex as second function and integrating by parts, we obtain
I= x2∫ex dx - ∫{\((\frac {d}{dx}x^2)\)∫exdx} dx
I = x2ex - ∫2x.exdx
I = x2ex-2∫x.ex dx
Again integrating by parts,we obtain
I =x2ex - 2[x.∫exdx - ∫{\((\frac {d}{dx}x)\).∫exdx} dx]
I = x2ex - 2[xex-∫exdx]
I = x2ex - 2[xex - ex]
I = x2ex - 2xex + 2ex + C
I = ex(x2 - 2x + 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,
