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.

Don't forget that your differential changes from $dx$ to $du$. If $u = x^2+6$, then $du = \frac{du}{dx}dx = 2xdx$ or $dx = \frac{du}{2x}$. Making this substitution gives the missing $\frac{1}{2}$.
To see how it all comes together, notice that
$$\int x(x^2+6)^{10} dx = \int xu^{10}\frac{du}{2x} = \frac{1}{2}\int u^{10}du.$$
Can you take it from here?
