# Integrate ? 3sqrt(4x+3) dx

Apr 3, 2018

$\int 3 \sqrt{4 x + 3} \mathrm{dx} = \frac{\left(4 x + 3\right) \sqrt{4 x + 3}}{2} + C$

#### Explanation:

.

$\int 3 \sqrt{4 x + 3} \mathrm{dx}$

Let $u = 4 x + 3 , \therefore \mathrm{du} = 4 \mathrm{dx} , \therefore \mathrm{dx} = \frac{\mathrm{du}}{4}$

Le's substitute:

$\int 3 \sqrt{u} \frac{\mathrm{du}}{4} = \frac{3}{4} \int {u}^{\frac{1}{2}} \mathrm{du} = \frac{3}{4} \left(\frac{2}{3}\right) {u}^{\frac{3}{2}} = \frac{1}{2} {u}^{\frac{3}{2}}$

Let's substitute back:

$\int 3 \sqrt{4 x + 3} \mathrm{dx} = \frac{1}{2} {\left(4 x + 3\right)}^{\frac{3}{2}} = \frac{1}{2} \sqrt{{\left(4 x + 3\right)}^{3}}$

$\int 3 \sqrt{4 x + 3} \mathrm{dx} = \frac{\left(4 x + 3\right) \sqrt{4 x + 3}}{2} + C$

Apr 3, 2018

$3 \int \sqrt{4 x + 3} \mathrm{dx}$

Let $\left(4 x + 3\right) = t$
$\implies 4 \mathrm{dx} = \mathrm{dt}$

$3 \int \sqrt{4 x + 3} \mathrm{dx} = 3 \int \sqrt{t} \frac{\mathrm{dt}}{4}$

$\implies \frac{3}{4} \int \sqrt{t} \mathrm{dt}$

$\implies \frac{3}{4} \frac{{t}^{\frac{3}{2}}}{\frac{3}{2}} + c$

$\implies \frac{3}{4} \times \frac{2}{3} \left({t}^{\frac{3}{2}}\right) + c$

$\implies \frac{1}{2} {\left(4 x + 3\right)}^{\frac{3}{2}} + c$