Q:

Evaluate $\int x \sqrt{x+7}\,dx$

Consider the integral 
$$\int x \sqrt{x+7} \,dx$$
Let $u=x+7$, so that $x=u-7$, $du=dx$. Then the integral becomes
$$\int(u-7)u^{1/2}du $$
$$=\int\left(u^{3/2}-7u^{1/2}\right) du, $$
which evaluates to
$$\frac{2}{5}(x+7)^{5/2}-\frac{14}{3}(x+7)^{3/2}+C.$$
Now even though this is correct, the answer that my professor is looking for is 
$$\frac{2}{15}(x+7)^{3/2}(3x-14)+C $$
So my question is how do I manipulate my answer to get the one above?

A:

Simply observe that $(x+7)^{3/2}$ is "common throughout":
\begin{align*}
 \frac{2}{5}(x+7)^{5/2}-\frac{14}{3}(x+7)^{3/2} &= (x+7)^{3/2}\left[\frac25(x+7)^1-\frac{14}3\cdot 1\right]\\[3pt]
&= \frac2{15}(x+7)^{3/2}(3x-14).
\end{align*}
Another easy illustration of this is, for example, $x^{1/2} + x^{5/2} = x^{1/2}(1+x^2)$.

