Integrate $\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx$ I'm trying to solve
$$
\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx
$$
using the substitution rule.  The substitution rule, as far as I know it, reads

Let $g^\prime$ be a continuous function on $[a,b]$ whose range is an interval $I$, and let $f$ be continuous on $I$.  Then
  $$
\int_a^b f(g(x))g^\prime(x)\, dx = \int_{g(a)}^{g(b)} f(u)\, du
$$

The solutions manual gives the solution as follows.  Let $u = 16 + 5x$, so $x = \frac{1}{5}(u - 16)$ and $du = 5dx$.  When $x = 0$, $u = 16$ and when $x = 4$, $u = 36$. So,
\begin{align*}
\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx & = \int_{16}^{36} \frac{\frac{1}{5}(u - 16)}{\sqrt{u}} \, \frac{du}{5} = \ldots
\end{align*}
I'm comfortable solving from there, but I don't see exactly how I used the substitution rule as stated above.  It seems like we are saying $g(x) = 16 + 5x$, so $g^\prime(x) = 5$, and $f(u) = u^{-1/2}$.  Then $x = \frac{1}{5}(g(x) - 16)$, and we have the form
$$
\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx = \int_{0}^{4}\frac{1}{5}(g(x) - 16)f(g(x))\, dx,
$$
which is not what we need for the substitution rule.  How can I write the original integral in terms of $f$, $g$ and $g^\prime$ so it is apparent how to apply the substitution rule?
 A: $$\int_{0}^{4}\frac{x}{\sqrt{16+5x}}\space\text{d}x=$$

Substitute $u=16+5x$ and $\text{d}u=5\space\text{d}x$.
This gives a new lower bound $u=16+5\cdot0=16$ and upper bound $u=16+5\cdot4=36$:

$$\frac{1}{5}\int_{16}^{36}\frac{u-16}{5\sqrt{u}}\space\text{d}u=$$
$$\frac{1}{5}\int_{16}^{36}\left[\frac{\sqrt{u}}{5}-\frac{16}{5\sqrt{u}}\right]\space\text{d}u=$$
$$\frac{1}{5}\left[\int_{16}^{36}\frac{\sqrt{u}}{5}\space\text{d}u-\int_{16}^{36}\frac{16}{5\sqrt{u}}\space\text{d}u\right]=$$
$$\frac{1}{5}\left[\frac{1}{5}\int_{16}^{36}\sqrt{u}\space\text{d}u-\frac{16}{5}\int_{16}^{36}\frac{1}{\sqrt{u}}\space\text{d}u\right]=$$
$$\frac{1}{5}\left[\frac{1}{5}\int_{16}^{36}u^{\frac{1}{2}}\space\text{d}u-\frac{16}{5}\int_{16}^{36}u^{-\frac{1}{2}}\space\text{d}u\right]=$$

Use:
$$\int y^b\space\text{d}y=\frac{y^{b+1}}{b+1}+\text{C}$$

$$\frac{1}{5}\left[\frac{1}{5}\left[\frac{2u^{\frac{3}{2}}}{3}\right]_{16}{36}-\frac{16}{5}\left[2\sqrt{u}\right]_{16}{36}\right]=$$
$$\int y^b\space\text{d}y=\frac{y^{b+1}}{b+1}+\text{C}$$

$$\frac{1}{5}\left[\frac{2}{15}\left[u^{\frac{3}{2}}\right]_{16}{36}-\frac{32}{5}\left[\sqrt{u}\right]_{16}{36}\right]=$$
$$\frac{1}{5}\left[\frac{2}{15}\left(36^{\frac{3}{2}}-16^{\frac{3}{2}}\right)-\frac{32}{5}\left(\sqrt{36}-\sqrt{16}\right)\right]=$$
$$\frac{1}{5}\left[\frac{2}{15}\left(216-64\right)-\frac{32}{5}\left(6-4\right)\right]=$$
$$\frac{1}{5}\left[\frac{2}{15}\left(152\right)-\frac{32}{5}\left(2\right)\right]=$$
$$\frac{1}{5}\left[\frac{304}{15}-\frac{64}{5}\right]=$$
$$\frac{1}{5}\left[\frac{112}{15}\right]=$$
$$\frac{112}{75}$$
A: Let $g(x)=16+5x$. Then, $x=\frac15(g(x)-16)$.
$$\int_0^4 \frac{x}{\sqrt{16 + 5x}} dx=\int_0^4\frac{\frac15(g(x)-16)}{\sqrt{g(x)}}dx$$
Then, $$f(x)=\frac{\frac15(x-16)}{\sqrt{x}}$$
This gives,
$$\int_0^4 \frac{x}{\sqrt{16 + 5x}} dx=\int_0^4f(g(x))g'(x)dx$$
