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. 
 A: Because we have,$$u = 4x +1 \implies u - 1 = 4x \implies \frac{u - 1}{4} = x$$
Can you do the rest?
A: 
$$I= \int{x\sqrt{4x+1}}dx$$

Using Euler Substitution
$$\sqrt{4x+1}=t\iff x=\frac{t^2-1}{4}\iff dx=\frac{t}{2}dt$$
$$\begin{align}
I&= \int{x\sqrt{4x+1}}dx\\
&=\int \left(\frac{t^2-1}{4}\right)\frac{t^2}{2}dt\\
&=\frac{t^5}{40}-\frac{t^3}{24}+C\\
&I=\frac{(4x+1)^{5/2}}{40}-\frac{(4x+1)^{3/2}}{24}+C\\
\end{align}$$

$$\int{x\sqrt{4x+1}}dx=\frac{3(4x+1)^{5/2}-5(4x+1)^{3/2}}{120}+C$$

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

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

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\begin{align}&\color{#66f}{\large\int x\root{4x + 1}\,\dd x}
={1 \over 4}\bracks{%
\int\pars{4x + 1}^{3/2}\,\dd x - \int\pars{4x + 1}^{1/2}\,\dd x}
\\[3mm]&={1 \over 4}\bracks{{\pars{4x + 1}^{5/2} \over 4\times 5/2}
-{\pars{4x + 1}^{3/2} \over 4\times 3/2}}
=\color{#66f}{\large{3\pars{4x + 1}^{5/2} - 5\pars{4x + 1}^{3/2} \over 120}}
+ \pars{~\mbox{a constant}~}
\end{align}
