What is the antiderivative of sqrt(x+3)?

May 2, 2018

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

Explanation:

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

$\textcolor{g r e e n}{\int {\left(x + a\right)}^{n} \mathrm{dx} = {\left(x + a\right)}^{n + 1} / \left(n + 1\right) + C}$

intsqrt(x+3)dx=color(blue)((x+3)^(1/2+1)/(1/2+1)+C

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

May 2, 2018

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

Explanation:

Finding the antiderivative of a function is the same as finding its integral (by the Fundamental Theorem of Calculus).

To find $\int \sqrt{x + 3} \mathrm{dx}$, we can use recognition or a natural substitution. We will use the latter.

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

$\int \sqrt{x + 3} \mathrm{dx} = \int \sqrt{u} \mathrm{du} = \int {u}^{\frac{1}{2}} \mathrm{du}$

Now we employ the power rule for integration:

$\int {x}^{n} \mathrm{dx} = \frac{1}{n + 1} {x}^{n + 1} + \text{c}$

Thus

$\int {u}^{\frac{1}{2}} \mathrm{du} = \frac{2}{3} {u}^{\frac{3}{2}} + \text{c"=2/3(x+3)^(3/2)+"c}$

May 2, 2018

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

Explanation:

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