A question regarding integrating Consider the function $ f(x) = (2x-1)^6$. I want to find the integral. Intuitively to me, it would be $F(x) = \dfrac{1}{7} (2x-1)^7 + c$, but for some reason you have to add $\dfrac{1}{a} = \dfrac{1}{2}$ too. I have never really understood why. Why?
 A: Think of inverting this via the Chain rule. Note that
$$
\frac{dF}{dx}
  = \frac{d}{dx} \frac{1}{7} (2x-1)^7
  = \frac{1}{7} 7 (2x-1)^6 \frac{d(2x-1)}{dx} 
  = \frac{1}{7} 7 (2x-1)^6 \cdot 2. 
$$
That last $2$ needs to be counterbalanced in integration, that's where the factor of $1/2$ comes from.
A: Correction: you need to "multiply" the integral by $\dfrac 12$.
Try differentiating $\quad \dfrac 17(2x- 1)^7 + C$: 
$$\dfrac d{dx}\left(\frac 17(2x - 1\right)^7 + C = (2x - 7)^6 \cdot \frac d{dx}\left(2x - 7\right) = 2(2x - 1)^6$$
You don't end up with the original integrand, because we need to use the chain rule to differentiate.
To this end, you can think of integrating $(2x - 1)^6$ by letting $\color{red}{\bf u = 2x - 1}$, $du = 2\,dx \implies \color{blue}{\bf dx = \frac 12 \,du}$. Then 
$$\int \color{red}{\bf (2x - 1)}{\bf ^6} \,\color{blue}{\bf dx} = \int (\color{red}{\bf u}^{\bf 6}) \color{blue}{\bf \frac 12 du} = {\bf \dfrac{1}{2} \int u^6\,du}$$
A: Differentiate your answer. You will see that an extra 2 appears in the denominator as per chain rule. So you need to divide by two so as to get the same function as the derivative. 
A: The answer concerns the chain rule.  Notice that $F'(x) = (2x-1)^6 \cdot \frac{d}{dx} (2x-1)$.
In general, you cannot integrate $g(x)^6$ by treating $g(x)$ like $x$.  You can, however, integrate $\int g(x)^6 g'(x) dx = \frac{1}{7} g(x)^7$.  This is the basis of the technique of $u$-substitution.
Applying $u$-substitution to this problem, we would set $u = 2x-1$.  Then, $du = 2dx$.
$$\int (2x-1)^6 dx = \frac{1}{2} \int u^6 du = \frac{1}{14} u^7 + C = \frac{1}{14} (2x-1)^7 + C$$
