1. Can someone please explain the steps in this integral problem?

∫x(x^2+1)^5 dx from 0 to 2

Part I know:
substitution is u=x^2+1, du=2x dx, x=0 u=1, x=2 u=5

Part from solutions manual:
(1/2) ∫ (x^2+1)^5 (2x dx)
(1/2) ∫ u^5 du from x=1 to 5
[u/12] from 1 to 5
(5^6-1^6)/12
15624/12

Now my questions:
Where did the 1/2 in front of the integral come from?
Where did the u/12 come from?
Is there a trick to this that I am just not seeing? -- Calc 2 is kicking my butt so far.

2. Originally Posted by operaphantom2003
∫x(x^2+1)^5 dx from 0 to 2

Part I know:
substitution is u=x^2+1, du=2x dx, x=0 u=1, x=2 u=5

Part from solutions manual:
(1/2) ∫ (x^2+1)^5 (2x dx)
(1/2) ∫ u^5 du from x=1 to 5
[u/12] from 1 to 5
(5^6-1^6)/12
15624/12

Now my questions:
Where did the 1/2 in front of the integral come from?
Where did the u/12 come from?
Is there a trick to this that I am just not seeing? -- Calc 2 is kicking my butt so far.
$\displaystyle \int x(x^2+1)^5 \, dx$

if $\displaystyle u = x^2+5$, then $\displaystyle du = 2x \, dx$

multiply the integral by $\displaystyle \frac{1}{2} \cdot 2$ ... the $\displaystyle \frac{1}{2}$ outside the integral, the "$\displaystyle 2$" inside to give you $\displaystyle du$

$\displaystyle \frac{1}{2} \int 2x(x^2+1)^5 \, dx$

now substitute ...

$\displaystyle \frac{1}{2} \int u^5 \, du$

antiderivative is ...

$\displaystyle \frac{1}{2} \cdot \frac{u^6}{6} + C$

multiply the fractions ... see where the $\displaystyle \frac{u}{12}$ comes from?