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Powers, roots and rational functions

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.

Powers, roots and rational functions Updated 2 min read

At a glance

\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C
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}
On this page
  1. Graph of √x and the area under it
  2. Derivation, step by step
  3. A second derivation: area by slicing sideways
  4. Check by differentiating
  5. Other roots and powers of x
  6. Substitution variant: integral of √(ax + b)
  7. Definite integral examples
  8. Common mistakes

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

Area under y = √x from 0 to 4, with the antiderivative y = (2/3)x^(3/2)
The shaded area is \frac{16}{3} \approx 5.33. The dashed antiderivative reaches \frac{16}{3} at x = 4.

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 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. 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 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.

Practise on antiderivative problems.

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The reverse power rule

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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.

Written by the PracticeCalculus team

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Related integrals

Integral of x \int x\,dx = \frac{x^2}{2} + C The integral of x is x²/2 + C. See why with the reverse power rule and a triangle, plus definite integrals, the linear form ax + b and common mistakes. Integral of x² \int x^2\,dx = \frac{x^3}{3} + C The integral of x² is x³/3 + C. See the reverse power rule, a Riemann sum proof, definite integrals, the (ax + b)² form and the mistakes students make. Integral of 1/x² \int \frac{1}{x^2}\,dx = -\frac{1}{x} + C The integral of 1/x² is -1/x + C. Rewrite it as x^(-2), check by differentiating, and see improper integrals like 1 to infinity and the classic -1 to 1 trap.