# Integral of √x: Formula, Derivation and Examples

> The integral of √x is (2/3)x^(3/2) + C. Rewrite the root as a power, check by differentiating, and see √(ax + b), 1/√x and definite integral examples.

Source: https://practicecalculus.com/integrals/sqrt-x
Updated: 2026-11-06

$$ \int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C $$

## At a glance

- **Antiderivative:** $\dfrac{2}{3}x^{3/2} + C = \dfrac{2}{3}x\sqrt{x} + C$
- **Rewrite first:** $\sqrt{x} = x^{1/2}$
- **Domain:** $x \ge 0$
- **Related integral:** $\int \dfrac{1}{\sqrt{x}}\,dx = 2\sqrt{x} + C$
- **Check:** $\dfrac{d}{dx}\dfrac{2}{3}x^{3/2} = x^{1/2}$

## Key takeaways

- The integral of $\sqrt{x}$ is $\frac{2}{3}x^{3/2} + C$.
- Always rewrite roots as fractional powers, $\sqrt{x} = x^{1/2}$, before using the reverse power rule.
- Dividing by $\frac{3}{2}$ is the same as multiplying by $\frac{2}{3}$.
- The integral of $\frac{1}{\sqrt{x}}$ is $2\sqrt{x} + C$, a result worth remembering.

The **integral of $\sqrt{x}$ is $\dfrac{2}{3}x^{3/2} + C$**. The trick is to write the square root as a power first.

$$\int \sqrt{x}\,dx = \int x^{1/2}\,dx = \frac{2}{3}x^{3/2} + C$$

## Graph of √x and the area under it

```graph
title: 'Area under y = √x from 0 to 4, with the antiderivative y = (2/3)x^(3/2)'
caption: 'The shaded area is $\frac{16}{3} \approx 5.33$. The dashed antiderivative reaches $\frac{16}{3}$ at $x = 4$.'
x: [-0.5, 5]
y: [-0.5, 7.5]
functions:
  - expr: sqrt(x)
    label: y = √x
    domain: [0, 5]
  - expr: (2/3)x^(3/2)
    label: y = (2/3)x^(3/2) (antiderivative)
    dashed: true
    domain: [0, 5]
shade: {expr: sqrt(x), from: 0, to: 4}
points:
  - [4, 5.333, '(4, 16/3)']
```

The shaded region fits inside a $4$ by $2$ rectangle of area $8$. It fills exactly two thirds of it, which is where the $\frac{2}{3}$ comes from.

## Derivation, step by step

1. Rewrite the root: $\sqrt{x} = x^{1/2}$.
2. Use the [reverse power rule](/rules/reverse-power-rule) with $n = \frac{1}{2}$.
3. Add one to the exponent: $\frac{1}{2} + 1 = \frac{3}{2}$.
4. Divide by $\frac{3}{2}$, which means multiply by $\frac{2}{3}$: $\frac{2}{3}x^{3/2}$.
5. Add $C$.

## A second derivation: area by slicing sideways

There is a neat way to see the $\frac{2}{3}$ without any rule. The curve $y = \sqrt{x}$ is the same as $x = y^2$ for $y \ge 0$. Draw the rectangle from $(0, 0)$ to $(b, \sqrt{b})$. The curve splits it in two. The piece between the curve and the $y$-axis is the area under $x = y^2$ from $y = 0$ to $y = \sqrt{b}$, which is $\frac{(\sqrt{b})^3}{3}$ by the [integral of x²](/integrals/x-squared). The full rectangle is $b\sqrt{b}$, so the area under $\sqrt{x}$ is what is left over:

$$b\sqrt{b} - \frac{b\sqrt{b}}{3} = \frac{2}{3}b^{3/2}$$

That matches the formula. It also shows why the two areas split in the ratio $1 : 2$.

## Check by differentiating

$$\frac{d}{dx}\left(\frac{2}{3}x^{3/2}\right) = \frac{2}{3}\cdot\frac{3}{2}x^{1/2} = \sqrt{x} \checkmark$$

The forward rule is on the [derivative of √x](/derivatives/sqrt-x) page.

## Other roots and powers of x

The same method works for any root. Rewrite, then raise the power and divide:

| Integrand | As a power | Integral |
| --- | --- | --- |
| $\sqrt{x}$ | $x^{1/2}$ | $\frac{2}{3}x^{3/2} + C$ |
| $\frac{1}{\sqrt{x}}$ | $x^{-1/2}$ | $2x^{1/2} + C$ |
| $x\sqrt{x}$ | $x^{3/2}$ | $\frac{2}{5}x^{5/2} + C$ |
| $\sqrt[3]{x}$ | $x^{1/3}$ | $\frac{3}{4}x^{4/3} + C$ |
| $\frac{x + 1}{\sqrt{x}}$ | $x^{1/2} + x^{-1/2}$ | $\frac{2}{3}x^{3/2} + 2x^{1/2} + C$ |

## Substitution variant: integral of √(ax + b)

Let $u = ax + b$, so $du = a\,dx$:

$$\int \sqrt{ax + b}\,dx = \frac{1}{a}\int u^{1/2}\,du = \frac{2}{3a}(ax + b)^{3/2} + C$$

For example, $\int \sqrt{3x + 2}\,dx = \frac{2}{9}(3x + 2)^{3/2} + C$. Check: $\frac{2}{9}\cdot\frac{3}{2}(3x+2)^{1/2}\cdot 3 = \sqrt{3x + 2}$.

A classic: $\int x\sqrt{x^2 + 1}\,dx = \frac{1}{3}(x^2 + 1)^{3/2} + C$, with $u = x^2 + 1$.

## Definite integral examples

**Example 1.** $\int_1^9 \sqrt{x}\,dx = \frac{2}{3}\left(9^{3/2} - 1^{3/2}\right) = \frac{2}{3}(27 - 1) = \frac{52}{3}$.

**Example 2.** $\int_0^1 \frac{1}{\sqrt{x}}\,dx = \Big[2\sqrt{x}\Big]_0^1 = 2$. This is an improper integral, because $\frac{1}{\sqrt{x}}$ blows up at $0$, but the area is still finite.

**Example 3.** $\int_0^4 (\sqrt{x} - 1)\,dx = \frac{16}{3} - 4 = \frac{4}{3}$.

## Common mistakes

1. **Using the power rule on the root sign directly.** Write $x^{1/2}$ first.
2. **Dividing by $\frac{1}{2}$ instead of $\frac{3}{2}$.** You divide by the new exponent, not the old one.
3. **Writing $\frac{3}{2}x^{3/2}$.** Dividing by $\frac{3}{2}$ gives $\frac{2}{3}$, not $\frac{3}{2}$.
4. **Integrating $\sqrt{x^2 + 1}$ like $\sqrt{x}$.** With a non-linear inside and no matching factor, substitution fails. That integral needs [trig substitution](/blog/trig-substitution).

Practise on [antiderivative problems](/practice-problems/antiderivatives).

## Frequently asked questions

### What is the integral of √x?

The integral of $\sqrt{x}$ is $\frac{2}{3}x^{3/2} + C$, which you can also write as $\frac{2}{3}x\sqrt{x} + C$.

### What is the integral of 1/√x?

It is $2\sqrt{x} + C$. Write $\frac{1}{\sqrt{x}} = x^{-1/2}$, raise the power to $\frac{1}{2}$ and divide by $\frac{1}{2}$.

### What is the integral of √x from 0 to 4?

It is $\frac{2}{3}\cdot 4^{3/2} = \frac{2}{3}\cdot 8 = \frac{16}{3}$.

### What is the integral of x√x?

Combine the powers first: $x\sqrt{x} = x^{3/2}$, so the integral is $\frac{2}{5}x^{5/2} + C$.