How do you evaluate the integral int sqrt(2x+3)dx?

Jan 9, 2017

$= \frac{1}{3} {\left(2 x + 3\right)}^{\frac{3}{2}} + C$

Explanation:

Let $u = 2 x + 3$. Then $\mathrm{du} = 2 \mathrm{dx}$ and $\mathrm{dx} = \frac{1}{2} \mathrm{du}$

$= \frac{1}{2} \int \sqrt{u} \mathrm{du}$

$= \frac{1}{2} \int {u}^{\frac{1}{2}} \mathrm{du}$

You can integrate this as $\int \left({x}^{n}\right) \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$:

$= \frac{1}{2} \left(\frac{2}{3} {u}^{\frac{3}{2}}\right) + C$

$= \frac{1}{3} {u}^{\frac{3}{2}} + C$

$= \frac{1}{3} {\left(2 x + 3\right)}^{\frac{3}{2}} + C$

Hopefully this helps!