Does $\int x^ndx = \frac{1}{n + 1}x^{n + 1} + c$ for all real $n$? I know it's established where $n$ is integer except for $-1$.
$$\int x^ndx = \frac{1}{n + 1}x^{n + 1} + c$$
But, I'm not sure if it's established where $n$ is real number.
Is it established where n is real number except for $-1$?
If so, would you introduce the proof please?
 A: If you have $y = x^\pi$ take logs and differentiate:
$$\ln y = \ln x^\pi = \pi \ln x$$
$$\frac{1}{y}y' = \pi \frac{1}{x}.$$
Multiply both sides by $y$ and then remember what $y$ was:
$$y' = \pi \frac{1}{x} y = \pi \frac{1}{x} x^\pi = \pi x^{\pi -1}.$$
So that's how the differentiation rule works for any real number.  And so your anti-derivative rule works for any real number except $n = -1.$ 
A: Are you happy with the corresponding rule for differentiation? If so then this follows immediately from the fundamental theorem of calculus.
Let $n \ne 1$ be an arbitrary real number and $f(x)=\frac{1}{n+1}x^{n+1}$.
FTC says that if $f$ is $C^1$ (which it is since f is a poly) then $$\int_a^x f'(t)dt = f(x)-f(a)$$ which exactly says $$\int_a^x t^ndt = \frac{1}{n+1}x^{n+1} - \frac{1}{n+1}a^{n+1}$$ or in indefinite integral notation
$$\int x^ndx = \frac{1}{n+1}x^{n+1} + c$$
To prove the derivative of $x^n$ for $n$ real, use Newton's generalised binomial theorem and the limit definition of the derivative.
