Atenção: Quando é válido integrar com expoentes não inteiros?

Happy Thanksgiving, reader! Let's have some fun applying the Power Rule for Integer Powers by doing problems 1, 2, 3, 4, 7, 11, 13, 14, 17, and 18 on this worksheet. Have a wonderful holiday!

Suppose we have the function $f(x)=x^n$, for any integer $n$. We would like to find the antiderivative of $f$.

Let us review some differentiation formulas; reversing these formulas will instantly give us what we want.

First, we will differentiate $f(x)=x^n$, for $n\ge0$: $$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}$$ $$\frac{d}{dx}x^n=\lim_{h\to0}\frac{(x+h)^n-x^n}{h}$$ $$\frac{d}{dx}x^n=\lim_{h\to0}\frac{\sum_{k=0}^{n}{n\choose k}x^kh^{n-k}-x^n}{h}$$ $$\frac{d}{dx}x^n=\lim_{h\to0}\sum_{k=0}^{n-1}{n\choose k}x^kh^{n-k-1}$$ $$\frac{d}{dx}x^n=\lim_{h\to0}(nx^{n-1}+\frac{n(n-1)}{2}x^{n-2}h+...)$$ $$\frac{d}{dx}x^n=nx^{n-1}$$ Now, we will differentiate $g(x)=\frac{1}{x^n}$, for $n\ge0$: $$\frac{d}{dx}\frac{1}{x^n}=-\frac{\frac{d}{dx}x^n}{(x^n)^2}$$ $$\frac{d}{dx}\frac{1}{x^n}=-\frac{nx^{n-1}}{x^{2n}}$$ $$\frac{d}{dx}x^{-n}=-nx^{-n-1}$$ In fact, if we replace $-n$ with $n$ in the above equation, we get: $$\frac{d}{dx}x^{n}=nx^{n-1}$$ This is actually exactly the same as for nonnegative integers!

Now, we are ready to integrate. Let $g(x)=f'(x)$. Then we have the following: $$f(x)=x^n$$ $$f'(x)=g(x)=nx^{n-1}$$ $$\int g(x)dx=f(x)+C$$ $$\int nx^{n-1}dx=x^n+C$$ Replacing $n-1$ with $n$ and dividing by $n$, we obtain the desired antiderivative: $$\int x^ndx=\frac{x^{n+1}}{n+1}+C$$ And now, we run into the only flaw: what if $n+1=0$?

Indeed, this is an issue. We must find something else for the antiderivative of $f(x)=x^{-1}$.

Our savior will just so happen to come in the form of the natural logarithm function.

Recall the following derivation of its derivative: $$g(x)=\ln x$$ $$x=e^{g(x)}$$ $$\frac{d}{dx}x=\frac{d}{dx}e^{g(x)}$$ $$1=e^{g(x)}\frac{d}{dx}g(x)$$ $$1=x\frac{d}{dx}g(x)$$ $$\frac{d}{dx}g(x)=\frac{1}{x}$$ The derivative of the logarithm function is the reciprocal function! Thus, we may say the following: $$\int \frac{dx}{x}=\ln x+C$$ But the logarithm function is only defined for $x\gt0$, so we do the following for maximal generality: $$\int \frac{dx}{x}=\ln|x|+C$$ Thus, we have our desired integration formulas: $$\int x^ndx=\frac{x^{n+1}}{n+1}+C\text{ }(n\ne-1)$$ $$\int \frac{dx}{x}=\ln|x|+C$$