# Indefinite Integral Formula

1. Jan 31, 2007

### 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?

Last edited: Jan 31, 2007
2. Feb 1, 2007

### dextercioby

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.

3. Feb 1, 2007

### HallsofIvy

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.