Power Rule for Indefinite Integrals To prove  $\int x^p \, dx = \frac{x^{p+1}}{p+1} + C$, my calculus textbook writes:

$$F '(x) = \frac{d}{dx} \left(\frac{x^{p+1}}{p+1} +C\right) = \frac{d}{dx} \left(\frac{x^{p+1}}{p+1}\right)+\frac{d}{dx}(C)=\frac{(p+1)x^p}{p+1}+0=x^p.$$

I am confused on how they take the derivative of $\dfrac{x^p+1}{p+1}$ without using the quotient rule. Can someone please explain to me why it is that they apply the power rule to the numerator but seem to completely ignore the denominator?
$p$ is a constant, so $\frac{1}{p+1}$ is a constant, so $\frac{x^{p+1}}{p+1}=\frac{1}{p+1}x^{p+1}$.  Just because there's a quotient doesn't mean we need to think of it as a quotient.
By definition,
$$
{d \over dx}x^n = \lim_{h \rightarrow 0} {(x+h)^n - x^n \over h}.
$$
Now, by Newton's binomial theorem,
$$
(x + h)^n = x^n + n h x^{n-1} + (\mbox{2nd and higher powers of $h$}).
$$
Subtracting $x^n$ and dividing by $h$, we get
$$
{(x+h)^n - x^n \over h} = n x^{n-1} + (\mbox{1st and higher powers of $h$})
$$
Now, letting $h \rightarrow 0$ on the right-hand side leaves just $nx^{n-1}$.