Using substitution to evaluate indefinite integral $ \int{x\sqrt{4x+1}}dx$

Question

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.

Answer 1

Because we have,$$u = 4x +1 \implies u - 1 = 4x \implies \frac{u - 1}{4} = x$$

Can you do the rest?

Answer 2

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

let $u$ = $4x+1$

$x = \frac{u-1}{4}$

$ \text{d}x = \frac{\text{d}u}{4}$

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