# Integration of cube root of x | Integral of cube root of x

The cube root of x is a number whose cube is x, in other words, (cube root of x)^3 = x. In this post, we will evaluate the integration of the cube root of x. To do so, we will use the power rule of derivatives which is given below:

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

where $c$ is an integration constant.

Also Read: Integration of root x

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

## Integration of Cube Root of x

Question: Find the integration of cube root x, that is, find

$\int \sqrt[3]{x} dx$

At first, by the rule of indices, we can write the cube root of $x$ as $x^{1/3}$. That is,

$\sqrt[3]{x}=x^{1/3}$

Using this fact, we have

$\int \sqrt[3]{x} dx$ $= \int x^{1/3} dx$

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

$=\dfrac{x^{1/3+1}}{4/3}+c$

$=\dfrac{3}{4} x^{1+1/3}+c$

$=\dfrac{3}{4} x \cdot x^{1/3}+c$

$=\dfrac{3}{4} x \sqrt[3]{x}+c$

So the integration of cube root of x is equal to $\dfrac{3}{4} x$ ∛x. In other words,

$\int$ ∛x $dx=\dfrac{3}{4} x$ ∛x +c where $c$ is an integration constant.

Derivative of x sin x

## Integration of 1 by Cube Root x

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

$\int \dfrac{1}{\sqrt[3]{x}} dx$

By the power rule of indices, we have  1/∛x $=1/x^{1/3}$. This can be further written as $x^{-1/3}$. Therefore, we have

$\int \dfrac{1}{\sqrt[3]{x}} dx$ $=\int x^{-1/3} dx$

Now applying the above power rule (i) of derivatives for $n=-1/3$, this is equal to

$= \int x^{-1/3} dx$

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

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

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

So the integration of 1 by cube root x is 3/2 x^{3/2}+c, where $c$ is an integration constant.

## Definite integration of Cube Root x

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

Question: Evaluate the integration of cube root x from 0 to 1.

$\int_0^1 \sqrt[3]{x} dx$

From the above we know that $\int \sqrt[3]{x} dx=\dfrac{3}{4} x \sqrt[3]{x}$.

So we have

$\int_0^1 \sqrt[3]{x} dx$ =$[\dfrac{3}{4} x \sqrt[3]{x}]_0^1$

$=\dfrac{3}{4}[x \sqrt[3]{x}]_0^1$

$=\dfrac{3}{4}[1-0]$

$=\dfrac{3}{4}$.