I am missing 1/2 in this integral I am trying to solve a integral for cal 2
using substitution.

$$\int x(6+x^2)^{10} dx , u = 6 + x^2$$

I know the answer is

$$\frac{(6+x^2)^{11}}{22}+C$$

i can't figure out where the last $\frac{1}{2}$ came from.

Observe carefully that if $u = 6 + x^2$, then $du = 2x\,dx$.  The integral becomes
$$\begin{aligned}
\int x(6 + x^2)^{10}\,dx &= \int \dfrac{1}{2}(2x\,dx)u^{10}\\
&= \dfrac{1}{2} \int u^{10}\,du\\
&= \dfrac{1}{2}\cdot \dfrac{u^{11}}{11} + \mbox{C}\\
&= \dfrac{(x^2 + 6)^{11}}{22} + \mbox{C}
\end{aligned}$$
$\frac{1}{2}$ comes from the substitution.  First, determine the differentials of the substitution.  Then, make the substitution of $x\,dx$ with the constant (If $du = 2x\,dx$, take $x\,dx$ and rewrite it as $2x\,dx$.  Then, multiply that by $\frac{1}{2}$ and substitute $2x\,dx$ with $du$).
Another way of seeing this is: Solve for $x\,dx$ and make the substitution.  This is equivalent to the first approach.
