Using substitution to evaluate indefinite integral $ \int{x\sqrt{4x+1}}dx$ 
Evaluate this indefinite integral.
  $$I= \int{x\sqrt{4x+1}}dx$$

Let $u=4x+1$
$$\frac{du}{dx}=4\rightarrow{dx=\frac{du}{4}}$$
$$I=\int{x}\sqrt{u}\frac{1}{4}du=\frac{1}{4}\int{x}\sqrt{u}du$$
Then I got stuck at this point. 

$$I = \int x \sqrt{4x+1} \ dx$$

let $u$ = $4x+1$
$x = \frac{u-1}{4}$
$ \text{d}x = \frac{\text{d}u}{4}$
$$
\begin{align}
I &= \int x \sqrt{4x+1} \ \text{d}x 
\\ &= \int \left(\frac{u-1}{4} \right) u^{1/2}  \frac{\text{d}u}{4}
\\ &= \frac{1}{16}\int u^{3/2}\text{d}u - \frac{1}{16}\int u^{1/2}\text{d}u
\\ &= \frac{1}{16}\frac{u^{5/2}}{5/2} - \frac{1}{16}\frac{u^{3/2}}{3/2} + \text{C}
\\ &= \frac{1}{40}u^{5/2} -  \frac{1}{24}u^{3/2} + \text{C}
\\ &= \frac{1}{40}(4x+1)^{5/2} - \frac{1}{24}(4x+1)^{3/2} + \text{C}
\end{align}
$$

