# Indefinite Integral Formula

• Izzhov

#### Izzhov

There is a simple formula for calculating $$\frac{df(x)}{dx} u^n$$ where u is a function of x and n is a positive rational number: $$\frac{df(x)}{dx} u^n = nu^{n-1} \ast \frac{du}{dx}$$. Is there a similar formula for calculating $$\int u^n dx$$ where u is a function of x and n is a positive rational number? It would be extremely helpful if there was.

P.S. I realize that the formula for $$\frac{df(x)}{dx} u^n$$ can be derived using the chain rule, so I was wondering if maybe the chain rule can somehow be applied in reverse for this problem?

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There is a simple formula for calculating $$\frac{df(x)}{dx} u^n$$ where u is a function of x and n is a positive rational number: $$\frac{df(x)}{dx} u^n = nu^{n-1} \ast \frac{du}{dx}$$. Is there a similar formula for calculating $$\int u^n dx$$ where u is a function of x and n is a positive rational number? It would be extremely helpful if there was.

P.S. I realize that the formula for $$\frac{df(x)}{dx} u^n$$ can be derived using the chain rule, so I was wondering if maybe the chain rule can somehow be applied in reverse for this problem?

Actually that "f(x)" should not be there. In the Leibniz notation the differentiation operator is simply $\frac{d}{dx}$.

As for the question itself, there's no general method for computing $\int u^{n}(x) {} dx$ for arbitrary "u(x)". In most cases, one can't find the antiderivative in terms of elemetary functions, even though "u(x)" may be elementary.

The "reverse" of the chain rule is basically integration by substitution. The problem is that when differentiating, using the chain rule, we are free to multiply by du/dx no matter what it is. When integrating, using substitution, that du/dx must already be in the integral- unless it is a constant, we can't just "put it in".

$$\int u^n \frac{du}{dx}dx= \int u^n du= \frac{1}{n+1}u^{n+1}$$
But that "du/dx" must already be in the integral.

The one exception to that is, as I said, when du/dx is a constant- when u is linear.
$$\int (ax+b)^n dx= \frac{1}{a(n+1)}(ax+b)^{n+1}$$
By letting u= ax+ b so that du= a dx and dx= (1/a)du.