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.
 A: 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?
A: 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.
A: Will it help to work the problem in reverse?  Let $y=\dfrac{(6+x^2)^{11}}{22}$.  Now we make the substitution $u=6+x^2$ and apply the chain rule.
$$y=\dfrac{u^{11}}{22},\frac{dy}{du}=\dfrac{u^{10}}2=\dfrac{(6+x^2)^{10}}2$$
$$u=6+x^2,\frac{du}{dx}=2x$$
$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\dfrac{2x(6+x^2)^{10}}2=x(6+x^2)^{10}$$
Now when integrating, the chain rule must still be considered.  We attempt to rewrite $\int f(x)dx$ as $\int g(u)du$.  If we wish to make the substitution
$$u=6+x^2,du=2xdx\text{, then }$$
$$\int x(6+x^2)^{10}dx=\int\frac12(6+x^2)^{10}(2xdx)=\frac12\int u^{10}du$$
To get from the first integral to the second integral, I just multiplied by $2$ and $\frac12$ (net result a multiplication by $1$) and rearranged terms to separate out $du$.  Then the constant was pulled outside the integral and the substitution applied.
