For my own use and consumption, I created a generalization of the nth integral of a function f and I'm posting it here for you look: $$\int f(x) dx = f^{(-1)}(x) + C_1$$ $$\iint f(x) dxdx = f^{(-2)}(x) + xC_1 + C_2$$ $$\iiint f(x) dxdxdx = f^{(-3)}(x) + \frac{1}{2}x^2C_1 + xC_2 + C_3$$(adsbygoogle = window.adsbygoogle || []).push({});

$$\int^n f(x) dx^n = f^{(-n)}(x) + \sigma_n(x)$$

being ##\sigma_n(x)##:

$$\sigma_0(x) = 0$$

$$\sigma_1(x) = \frac{1}{0!}x^0C_1$$

$$\sigma_2(x) = \frac{1}{1!}x^1C_1 + \frac{1}{0!}x^0C_2$$

$$\sigma_3(x) = \frac{1}{2!}x^2C_1 + \frac{1}{1!}x^1C_2 + \frac{1}{0!}x^0C_3$$

$$\sigma_4(x) = \frac{1}{3!}x^3C_1 + \frac{1}{2!}x^2C_2 + \frac{1}{1!}x^1C_3 + \frac{1}{0!}x^0C_4$$

Properties

$$\int^n \sigma_k(x) dx^n = \sigma_{k+n}(x)$$ $$\frac{d^n}{dx^n} \sigma_k(x) = \sigma_{k-n}(x)$$

Identities

$$f^{(-1)}(x) = \int_{x_0}^{x} f(x) dx + \frac{1}{0!}f^{(-1)}(x_0)(x-x_0)^0$$

$$f^{(-2)}(x) = \int_{x_0}^{x} \int_{x_0}^{x} f(x) dx dx + \frac{1}{1!}f^{(-1)}(x_0)(x - x_0)^1 + \frac{1}{0!}f^{(-2)}(x_0)(x-x_0)^0$$

$$f^{(-3)}(x) = \int_{x_0}^{x} \int_{x_0}^{x} \int_{x_0}^{x} f(x) dx dx dx + \frac{1}{2!}f^{(-1)}(x_0)(x - x_0)^2 + \frac{1}{1!}f^{(-2)}(x_0)(x - x_0)^1$$

$$ + \frac{1}{0!}f^{(-3)}(x_0)(x-x_0)^0$$

$$\int\limits_{x_0}^{x} \! {\;}^{n} f(x) dx^n = \frac{1}{(n-1)!} \int_{x_0}^{x} (x-u)^{(n-1)} f(u)du$$

What you think about, rox?

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# General formula for the nth integral

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