The integration of root x is equal to 2/3 x^{3/2}+C. In this post, we will calculate the integration of the square root of x. Note that the square root of x is an algebraic function. We will use the following power rule of derivatives:

$\int x^n dx = \dfrac{x^{n+1}}{n+1}+c$ $ \cdots (i)$

where $c$ is an integration constant.

**Note:** For more details of square roots, please click on the page Square Root of x: Definition, Symbol, Graph, Properties, Derivative, integration.

What is the integration of root x? Let’s find out.

## Integration of Root x

**Question:** Find the integration of root x, that is,

find $\int \sqrt{x} dx$

*Answer: *

At first, we will write the square root of x as $x^{\frac{1}{2}}$ by the rule of indices. That is,

$\sqrt{x}=x^{\frac{1}{2}}$

Using this fact, we have

$\int \sqrt{x} dx = \int x^{\frac{1}{2}} dx$

$=\dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c \quad$ by the above power rule (i) of derivatives.

$=\dfrac{x^{\frac{1}{2}+1}}{3/2}+c$

$=\dfrac{2}{3} x^{1+\frac{1}{2}}+c$

$=\dfrac{2}{3} x \cdot x^{\frac{1}{2}}+c$

$=\dfrac{2}{3} x\sqrt{x}+c$

So the integration of root x is equal to $\dfrac{2}{3} x\sqrt{x}$. In other words,

$\int \sqrt{x} dx =\dfrac{2}{3} x\sqrt{x}+c$ where $c$ is a constant.

**Also Read:** Derivative of √x | Derivative of 1/√x

Integration of $\sqrt{x}+\frac{1}{\sqrt{x}}$

## Integration of 1 by root x

**Question:** Find the integration of 1 by root x, that is, calculate

$\int \dfrac{1}{\sqrt{x}} dx$

*Answer:*

As $\sqrt{x}=x^{\frac{1}{2}}$ we have

$\dfrac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$

Using this fact, we have

$\int \dfrac{1}{\sqrt{x}} dx = \int x^{-\frac{1}{2}} dx$

$=\dfrac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c \quad$ by the above power rule (i) of derivatives.

$=\dfrac{x^{\frac{1}{2}}}{\frac{1}{2}}+c$

$=2\sqrt{x}+c$

So the integration of 1 by root x is 2√x+c, where c is an integration constant.

**Also Read: **Integration of cube root of x | Derivative of 1/(1+x^{2})

## Integration of x Root x

**Question:** What is the integration of x root x?

Evaluate $\int x \sqrt{x} dx$

*Answer:*

By the rule of indices, we have

$x\sqrt{x}=x^{1+\dfrac{1}{2}}=x^{3/2}$

Thus, we get that

$\int x\sqrt{x} dx = \int x^{3/2} dx$

$=\dfrac{x^{3/2+1}}{3/2+1}+c \quad$ by the above power rule (i) of derivatives.

$=\dfrac{x^{5/2}}{5/2}+c$

$=\dfrac{2}{5} x^{5/2}+c$

So the integration of x root(x) is equal to $2/5 x^{5/2}+c$. This means that

$\int x\sqrt{x} dx =\dfrac{2}{5} x^{5/2}+c$

**More Reading:** Integral of sin^{3}x | Find ∫sin^{3}x cos^{2}x dx

## Definite integration of Root x

Now, we will find a definite integral involving the square root of x.

**Question:** Evaluate $\int_0^1 \sqrt{x} dx$

*Answer:*

From above we know that $\int \sqrt{x} dx =\dfrac{2}{3} x\sqrt{x}$

So we have

$\int_0^1 \sqrt{x} dx=\dfrac{2}{3} [x\sqrt{x}]_0^1$

$=\dfrac{2}{3}[1\sqrt{1}-0\sqrt{0}]$

$=\dfrac{2}{3}[1-0]$

$=\dfrac{2}{3}$

So the definite integral of root x from 0 to 1 is equal to 2/3.

## FAQs

**Q1: What is the integration of root x?**

Answer: The integration of root x is 2/3 x^{3/2}+C where C is an integration constant.